Start Lecture #1

Chapter 0: Administrivia

I start at Chapter 0 so that when we get to chapter 1, the numbering will agree with the text.

0.1: Contact Information

• <my-last-name> AT nyu DOT edu (best method)
• http://cs.nyu.edu/~gottlieb
• 715 Broadway, Room 712
• 212 998 3344

0.2: Course Web Page

There is a web site for the course. You can find it from my home page listed above.

• You can also find these lecture notes on the course home page. Please let me know if you can't find it.
• The notes are updated as bugs are found or improvements made.
• I will also produce a separate page for each lecture after the lecture is given. These individual pages might not get updated as quickly as the large page.

0.3: Textbook

The course text is Aho, Lam, Sethi, and Ullman: Compilers: Principles, Techniques, and Tools, second edition

• Available in bookstore.
• We will cover most of the first 8 chapters (plus some asides).
• The first edition is a descendant of the classic Principles of Compiler Design.
• Independent of the titles, each of the books is called The Dragon Book, due to the cover picture.

0.4: Computer Accounts and Mailman Mailing List

• You are entitled to a computer account on one of the departmental sun machines. If you do not have one already, please get it asap.
• Sign up for the Mailman mailing list for the course. You can do so by clicking here
• If you want to send mail just to me, use the address given above, not the mailing list.
• Questions on the labs should go to the mailing list. You may answer questions posed on the list as well. Note that replies are sent to the list.
• I will respond to all questions; if another student has answered the question before I get to it, I will confirm if the answer given is correct.
• Please use proper mailing list etiquette.
  • Send plain text messages rather than (or at least in addition to) html.
  • Use the Reply command to contribute to the current thread, but NOT to start another topic.
  • If quoting a previous message, trim off irrelevant parts.
  • Use a descriptive Subject: field when starting a new topic.
  • Do not use one message to ask two unrelated questions.
  • Do NOT make the mistake of sending your completed lab assignment to the mailing list. This is not a joke; several students have made this mistake in past semesters.

0.5: Grades

Your grade will be a function of your final exam and laboratory assignments (see below). I am not yet sure of the exact weightings for each lab and the final, but the final will be roughly half the grade (very likely between 40% and 60%).

0.6: The Upper Left Board

I use the upper left board for lab/homework assignments and announcements. I should never erase that board. If you see me start to erase an announcement, please let me know.

I try very hard to remember to write all announcements on the upper left board and I am normally successful. If, during class, you see that I have forgotten to record something, please let me know. HOWEVER, if I forgot and no one reminds me, the assignment has still been given.

0.7: Homeworks and Labs

I make a distinction between homeworks and labs.

Labs are

• Required.
• Due several lectures later (date given on assignment).
• Graded and form part of your final grade.
• Penalized for lateness.
• Most often are computer programs you must write.

Homeworks are

• Optional.
• Due the beginning of the Next lecture.
• Not accepted late.
• Mostly from the book.
• Collected and returned.
• Able to help, but not hurt, your grade.

0.7.1: Homework Numbering

Homeworks are numbered by the class in which they are assigned. So any homework given today is homework #1. Even if I do not give homework today, the homework assigned next class will be homework #2. Unless I explicitly state otherwise, all homeworks assignments can be found in the class notes. So the homework present in the notes for lecture #n is homework #n (even if I inadvertently forgot to write it to the upper left board).

0.7.2: Doing Labs on non-NYU Systems

You may solve lab assignments on any system you wish, but ...

• You are responsible for any non-nyu machine. I extend deadlines if the nyu machines are down, not if yours are.
• Be sure test your assignments to the nyu systems. In an ideal world, a program written in a high level language like Java, C, or C++ that works on your system would also work on the NYU system used by the grader. Sadly this ideal is not always achieved despite marketing claims to the contrary. So, although you may develop you lab on any system, you must ensure that it runs on the nyu system assigned to the course.
• If somehow your assignment is misplaced by me and/or a grader, we need a to have a copy ON AN NYU SYSTEM that can be used to verify the date the lab was completed.
• When you complete a lab and have it on an nyu system, email the lab to the grader and copy yourself on that message. This email must come from your CIMS account. Keep the copy until you have received your grade on the assignment. The systems support staff can retrieve your mail from their logs given your copy and from that we can verify the dates. I realize that I am being paranoid about this. It is rare for labs to get misplaced, but they sometimes do and I really don't want to be in the middle of an I sent it ... I never received it debate. Thank you.

0.7.3: Obtaining Help with the Labs

Good methods for obtaining help include

1. Asking me during office hours (see web page for my hours).
2. Asking the mailing list.
3. Asking another student, but ...
4. ... Your lab must be your own.
   That is, each student must submit a unique lab. Naturally, simply changing comments, variable names, etc. does not produce a unique lab. See the Academic Integrity Policy below.

0.7.4: Computer Language Used for Labs

You may write your lab in Java, C, or C++. Other languages may be possible, but please ask in advance. I need to ensure that the TA is comfortable with the language.

0.8: A Grade of Incomplete

The rules for incompletes and grade changes are set by the school and not the department or individual faculty member. The rules set by GSAS state:

The assignment of the grade Incomplete Pass(IP) or Incomplete Fail(IF) is at the discretion of the instructor. If an incomplete grade is not changed to a permanent grade by the instructor within one year of the beginning of the course, Incomplete Pass(IP) lapses to No Credit(N), and Incomplete Fail(IF) lapses to Failure(F).

Permanent grades may not be changed unless the original grade resulted from a clerical error.

0.9: An Introductory Compiler Course with a Programming Prerequisite

0.9.1: An Introductory Course ...

I do not assume you have had a compiler course as an undergraduate, and I do not assume you have had experience developing/maintaining a compiler.

If you have already had a compiler class, this course is probably not appropriate. For example, if you can explain the following concepts/terms, the course is probably too elementary for you.

• Parsing
• Lexical Analysis
• Syntax analysis
• Register allocation
• LALR Grammar

0.9.2: ... with a Programming Prerequisite

I do assume you are an experienced programmer. There will be non-trivial programming assignments during this course. Indeed, you will write a compiler for a simple programming language.

I also assume that you have at least a passing familiarity with assembler language. In particular, your compiler may need to produce assembler language, but probably it will produce an intermediate language consisting of 3-address code. We will also be using addressing modes found in typical assemblers. We will not, however, write significant assembly-language programs.

0.10: Academic Integrity Policy

The CS policy on academic integrity, which applies to all graduate courses in the department, can be found here . A recent email from Rosemary Amico may also be of interest

    Dear faculty,

    The vast majority of our students comply with the
    department's academic integrity policies; see

      www.cs.nyu.edu/web/Academic/Undergrad/academic_integrity.html
      www.cs.nyu.edu/web/Academic/Graduate/academic_integrity.html

    Unfortunately, every semester we discover incidents in
    which students copy programming assignments from those of
    other students, making minor modifications so that the
    submitted programs are extremely similar but not identical.

    To help in identifying inappropriate similarities, we
    suggest that you and your TAs consider using Moss, a
    system that automatically determines similarities between
    programs in several languages, including C, C++, and Java.
    For more information about Moss, see:

      http://theory.stanford.edu/~aiken/moss/

    Feel free to tell your students in advance that you will be
    using this software or any other system.  And please emphasize,
    preferably in class, the importance of academic integrity.

    Rosemary Amico
    Assistant Director, Computer Science
  

An Interlude from Chapter 2

I present this snippet from chapter 2 here (it appears where it belongs as well), since it is self-contained and is needed for lab number 1, which I wish to assign today.

2.3.4: (depth-first) Tree Traversals

When performing a depth-first tree traversal, it is clear in what order the leaves are to be visited, namely left to right. In contrast there are several choices as to when to visit an interior (i.e. non-leaf) node. The traversal can visit an interior node

1. Before visiting any of its children.
2. Between visiting its children.
3. After visiting all of its children.

I do not like the book's pseudocode as I feel the names chosen confuse the traversal with visiting the nodes. I prefer the pseudocode below, which uses the following conventions.

• Comments are introduced by -- and terminate at the end of the line (as in the programming language Ada).
  
  
• Indenting is significant so begin/end or {} are not used (from the programming language family B2/ABC/Python)

    traverse (n : treeNode)
        if leaf(n)                      -- visit leaves once; base of recursion
           visit(n)
        else                            -- interior node, at least 1 child
           -- visit(n)                  -- visit node PRE visiting any children
           traverse(first child)        -- recursive call
           while (more children remain) -- excluding first child
               -- visit(n)              -- visit node IN-between visiting children
               traverse (next child)    -- recursive call
           -- visit(n)                  -- visit node POST visiting all children
  

Note the following properties

1. As written, with the last three visit()s commented out, only the leaves are visited and those visits are in left to right order.
   
   
2. If you uncomment just the first (interior node) visit, you get a preorder traversal, in which each node is visited before (i.e., pre) visiting any of its children.
   
   
3. If you uncomment just the last visit, you get a postorder traversal, in which each node is visited after (i.e., post) visiting all of its children.
   
   
4. If you uncomment only the middle visit, you get an inorder traversal, in which the node is visited (in-) between visiting its children.
   
   Inorder traversals are normally defined only for binary trees, i.e., trees in which every interior node has exactly two children. Although the code with only the middle visit uncommented works for any tree, we will, like everyone else, reserve the name inorder traversal for binary trees. In the case of binary search trees (everything in the left subtree is smaller than the root of that subtree, which in tern is smaller than everything in the corresponding right subtree) an inorder traversal visits the values of the nodes in (numerical) order.
   
   
5. If you uncomment two of the three visits, you get a traversal without a name.
   
   
6. If you uncomment all of the three visits, you get an Euler-tour traversal.

To explain the name Euler-tour traversal, recall that an Eulerian tour on a directed graph is one that traverses each edge once. If we view the tree on the right as undirected and replace each edge with two arcs, one in each direction, we see that the pink curve is indeed an Eulerian tour. It is easy to see that the curve visits the nodes in the order of the pseudocode (with all visits uncommented).

Normally, the Euler-tour traversal is defined only for a binary tree, but this time I will differ from convention and use the pseudocode above to define Euler-tour traversal for all trees.

Note the following points about our Euler-tour traversal.

• A node with k children is visited k+1 times. The diagram shows nodes with 0, 1, 2, and 3 children.
• In a binary tree, a leaf is visited once and an interior node is visited three times. This is one of the standard definitions of an Euler-tour traversal for a binary tree.
• The other standard definition has all nodes visited 3 times. For a leaf the three visits are in succession. Modifying the pseudocode to obtain this definition simply requires replacing the leaf visit with
      visit(n); visit(n); visit(n)

Do the Euler-tour traversal for the tree in the notes and then for a binary tree.

Lab 1 assigned. See the home page.

Roadmap of the Course

1. Chapter 1 touches on all the material.
   
   
2. Chapter 2 constructs (the front end of) a simple compiler.
   
   
3. Chapters 3-8 fill in the (considerable) gaps, as well as the presenting the beginnings of the compiler back end.
   
   

I always spend too much time on introductory chapters, but will try not to. I have said this before.

Chapter 1: Introduction to Compiling

Homework Read chapter 1.

1.1: Language Processors

A Compiler is a translator from one language, the input or source language, to another language, the output or target language.

Often, but not always, the target language is an assembler language or the machine language for a computer processor.

Note that using a compiler requires a two step process to run a program.

1. Execute the compiler (and possibly an assembler) to translate the source program into a machine language program.
2. Execute the resulting machine language program, supplying appropriate input.

This should be compared with an interpreter, which accepts the source language program and the appropriate input, and itself produces the program output.

Sometimes both compilation and interpretation are used. For example, consider typical Java implementations. The (Java) source code is translated (i.e., compiled) into bytecodes, the machine language for an idealized virtual machine, the Java Virtual Machine or JVM. Then an interpreter of the JVM (itself normally called a JVM) accepts the bytecodes and the appropriate input, and produces the output. This technique was quite popular in academia, with the Pascal programming language and P-code.

Homework: 1, 2, 4

Remark: Unless otherwise stated, homeworks are from the book and specifically from the end of the second level section we are discussing. Even more specifically, we are in section 1.1, so you are to do the first, second, and fourth problem at the end of section 1.1. These three problems are numbered 1.1.1, 1.1.2, and 1.1.4 in the book.

1.2: The Structure of a Compiler

Modern compilers contain two (large) parts, each of which is often subdivided. These two parts are the front end, shown in green on the right and the back end, shown in pink.

The front end analyzes the source program, determines its constituent parts, and constructs an intermediate representation of the program. Typically the front end is independent of the target language.

The back end synthesizes the target program from the intermediate representation produced by the front end. Typically the back end is independent of the source language.

This front/back division very much reduces the work for a compiling system that can handle several (N) source languages and several (M) target languages. Instead of NM compilers, we need N front ends and M back ends. For gcc (originally abbreviating Gnu C Compiler, but now abbreviating Gnu Compiler Collection), N=7 and M~30 so the savings are considerable.

Multiple Phases

The front and back end are themselves each divided into multiple phases. Conceptually, the input to each phase is the output of the previous. Sometime a phase changes the representation of the input. For example, the lexical analyzer converts a character stream input into a token stream output. Sometimes the representation is unchanged. For example, the machine-dependent optimizer transforms target-machine code into (hopefully improved) target-machine code.

The diagram is definitely not drawn to scale, in terms of effort or lines of code. In practice, the optimizers dominate.

Conceptually, there are three phases of analysis with the output of one phase the input of the next. Each of these phases changes the representation of the program being compiled. The phases are called lexical analysis or scanning, which transforms the program from a string of characters to a string of tokens; syntax analysis or parsing, which transforms the program into some kind of syntax tree; and semantic analysis, which decorates the tree with semantic information.

Note that the above classification is conceptual; in practice more efficient representations may be used. For example, instead of having all the information about the program in the tree, tree nodes may point to symbol table entries. Thus the information about the variable counter is stored once and pointed to at each occurrence.

1.2.1: Lexical Analysis (or Scanning)

The character stream input is grouped into meaningful units called lexemes, which are then mapped into tokens, the latter constituting the output of the lexical analyzer. For example, any one of the following C statements

    x3 = y + 3;
    x3  =   y   +   3   ;
    x3   =y+ 3  ;
  

    x 3 = y + 3;
  

A token is a <token-name,attribute-value> pair. For example

1. The lexeme x3 would be mapped to a token such as <id,1>. The name id is short for identifier. The value 1 is the index of the entry for x3 in the symbol table produced by the compiler. This table is used gather information about the identifiers and to pass this information to subsequent phases.
2. The lexeme = would be mapped to the token <=>. In reality it is probably mapped to a pair, whose second component is ignored. The point is that there are many different identifiers so we need the second component, but there is only one assignment symbol =.
3. The lexeme y is mapped to the token <id,2>
4. The lexeme + is mapped to the token <+>.
5. The lexeme 3 is somewhat interesting and is discussed further in subsequent chapters. It is mapped to <number,something>, but what is the something. On the one hand there is only one 3 so we could just use the token <number,3>. However, there can be a difference between how this should be printed (e.g., in an error message produced by subsequent phases) and how it should be stored (fixed vs. float vs. double). Perhaps the token should point to the symbol table where an entry for this kind of 3 is stored. Another possibility is to have a separate numbers table.
6. The lexeme ; is mapped to the token <;>.

Note that non-significant blanks are normally removed during scanning. In C, most blanks are non-significant. That does not mean the blanks are unnecessary. Consider

    int x;
    intx;
  

Note that we can define identifiers, numbers, and the various symbols and punctuation without using recursion (compare with parsing below).

1.2.2: Syntax Analysis (or Parsing)

Parsing involves a further grouping in which tokens are grouped into grammatical phrases, which are often represented in a parse tree. For example

    x3 = y + 3;
  

This parsing would result from a grammar containing rules such as

    asst-stmt → id = expr ;
    expr      → number
              |  id
              |  expr + expr
  

Note the recursive definition of expression (expr). Note also the hierarchical decomposition in the figure on the right.

The division between scanning and parsing is somewhat arbitrary, in that some tasks can be accomplished by either. However, if a recursive definition is involved, it is considered parsing not scanning.

Often one uses a simpler tree called the syntax tree (more properly the abstract syntax tree) with operators as interior nodes and operands as the children of the operator. The syntax tree on the right corresponds to the parse tree above it. We expand on this point later.

1.2.3: Semantic Analysis

There is more to a front end than simply syntax. The compiler needs semantic information, e.g., the types (integer, real, pointer to array of integers, etc) of the objects involved. This enables checking for semantic errors and inserting type conversion where necessary.

For example, if y was declared to be a real and x3 an integer, we need to insert (unary, i.e., one operand) conversion operators inttoreal and realtoint as shown on the right.

In this class we will use three-address-code for our intermediate language; another possibility that is used is some kind of syntax tree.

1.2.4: Intermediate code generation

Many compilers internally generate intermediate code for an idealized machine. For example, the intermediate code generated would assume that the target has an unlimited number of registers and that any register can be used for any operation. This is similar to a machine model with no registers, but which permits operations to be directly performed on memory locations. Another common assumption is that machine operations take (up to) three operands: two source and one target.

With these assumptions of a machine with an unlimited number of registers and instructions with three operands, one generates three-address code by walking the semantic tree. Our example C instruction would produce

    temp1 = inttoreal(3)
    temp2 = y + temp1
    temp3 = realtoint(temp2)
    x3 = temp3
  

We see that three-address code can include instructions with fewer than 3 operands.

Sometimes three-address code is called quadruples because one can view the previous code sequence as

    inttoreal temp1 3     --
    add       temp2 y     temp1
    realtoint temp3 temp2 --
    assign    x3    temp3 --
  

    operation  target source1 source2
  

1.2.5: Code optimization

This is a very serious subject, one that we will not really do justice to in this introductory course. Some optimizations, however, are fairly easy to understand.

1. Since 3 is a constant, the compiler can perform the int to real conversion and replace the first two quads with

   	add       temp2 y    3.0
         

2. The last two quads can be combined into

   	realtoint x3    temp2
         

In addition to optimizations performed on the intermediate code, further optimizations can be performed on the machine code by the machine-dependent back end.

1.2.6: Code generation

Modern processors have only a limited number of register. Although some processors, such as the x86, can perform operations directly on memory locations, we will for now assume only register operations. Some processors (e.g., the MIPS architecture) use three-address instructions. We follow this model. Other processors permit only two addresses; the result overwrites one of the sources. Using three-address instructions restricted to registers (except for load and store instructions, which naturally must also reference memory), code something like the following would be produced for our example, after first assigning memory locations to x3 and y.

    LD   R1,  y
    LD   R2,  #3.0        // Some would allow constant in ADD
    ADDF R1,  R1, R2      // add float
    RTOI R2,  R1          // real to int
    ST   x3,  R2
  

1.2.7: Symbol-Table Management

The symbol table stores information about program variables that will be used across phases. Typically, this includes type information and storage locations.

A possible point of confusion: the storage location does not give the location where the compiler has stored the variable. Instead, it gives the location where the compiled program will store the variable.

1.2.8: The Grouping of Phases into Passes

Logically each phase is viewed as a separate pass, i.e., a program that reads input and produces output for the next phase. The phases thus form a pipeline. In practice some phases are combined into a pass.

For example one could have the entire front end as one pass.

The term pass is used to indicate that the entire input is read during this activity. So two passes, means that the input is read twice. A grayed out (optional) portion of the notes above discusses 2-pass approaches for both assemblers and linkers. If we implement each phase separately and possibly use multiple passes for some of them, the compiler will perform a large number of I/O operations, an expensive undertaking.

As a result, techniques have been developed to reduce the number of passes. We will see in the next chapter how to combine the scanner, parser, and semantic analyzer into one program or phase. Consider the parser. When it needs to input the next token, rather than reading the input file (presumably produced by the scanner), the parser calls the scanner instead. At selected points during the production of the syntax tree, the parser calls the intermediate-code generator which performs semantic analysis as well as generating a portion of the intermediate code.

For pedagogical reasons, we will not be employing this technique. That is to ease the programming and understanding, we will use a compiler design that performs more I/O than necessary. Naturally, production compilers do not do this. Thus your compiler will consist of separate programs for the scanner, parser, and semantic analyzer / intermediate code generator. Indeed, these will likely be labs 2, 3, and 4.

1.2.9: Compiler-construction tools

Originally, compilers were written from scratch, but now the situation is quite different. A number of tools are available to ease the burden.

We will mention tools that generate scanners and parsers. This will involve us in some theory: regular expressions for scanners and context-free grammars for parsers. These techniques are fairly successful. One drawback can be that they do not execute as fast as hand-crafted scanners and parsers.

We will also see tools for syntax-directed translation and automatic code generation. The automation in these cases is not as complete.

Finally, there is the large area of optimization. This is not automated; however, a basic component of optimization is data-flow analysis (how values are transmitted between parts of a program) and there are tools to help with this task.

Pedagogically, a problem with using the tools is that your effort shifts from understanding how a compiler works to how the tool is used. So instead of studying regular expressions and finite state automata, you study the flex man pages and user's guide.

1.6: Programming Language Basics

Skipped. This is covered in our Programming Languages course, which is a prerequisite for the Compilers course.

Remark: You should be able to do the exercises in this section (but they are not assigned).

Chapter 2: A Simple Syntax-Directed Translator

Homework: Read chapter 2.

The goal of this chapter is to implement a very simple compiler. Really we are just going as far as the intermediate code, i.e., the front end. Nonetheless, the output, i.e. the intermediate code, does look somewhat like assembly language.

How is it possible to do this in just one chapter?
What is the rest of the book/course about?

In this chapter there is

1. A simple source language.
2. A target close to the source.
3. No optimization.
4. No machine-dependent back end.
5. No tools.
6. Little theory.

The material will be presented too fast for full understanding: Starting in chapter 3, we slow down and explain everything.

Sometimes in chapter 2, we only show some of the possibilities (typically omitting the hard cases) and don't even mention the omissions. Again, this is corrected in the remainder of the course.

A weakness of my teaching style is that I spend too long on chapters like this. I will try not to make that mistake this semester, but I have said that before.

2.1: Introduction

We will be looking at the front end, i.e., the analysis portion of a compiler.

The syntax describes the form of a program in a given language, while the semantics describes the meaning of that program. We will learn the standard representation for the syntax, namely context-free grammars also called BNF (Backus-Naur Form).

We will learn syntax-directed translation, where the grammar does more than specify the syntax. We augment the grammar with attributes and use this to guide the entire front end.

The front end discussed in this chapter has as source language infix expressions consisting of digits, +, and -. The target language is postfix expressions with the same components.

For example, the compiler will convert 7+4-5 to 74+5-. Actually, our simple compiler will handle a few other operators as well.

We will tokenize the input (i.e., write a scanner), model the syntax of the source, and let this syntax direct the translation all the way to three-address code, our intermediate language.

2.2: Syntax Definition

2.2.1: Definition of Grammars

This will be done right in the next two chapters.

A context-free grammar (CFG) consists of

1. A set of terminals (tokens produced by the lexer).
2. A set of nonterminals.
3. A set of productions (rules for transforming nonterminals). These are written
       LHS → RHS
   where the LHS is a single nonterminal (that is why this grammar is context-free) and the RHS is a string containing nonterminals and/or terminals.
4. A specific nonterminal designated as start symbol.

Example:

    Terminals: 0 1 2 3 4 5 6 7 8 9 + -
    Nonterminals: list digit
    Productions:
        list → list + digit
        list → list - digit
        list → digit
        digit → 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
    Start symbol: list
  

We use | to indicate that a nonterminal has multiple possible right hand side. So

    A → B | C
  

    A → B
    A → C
  

If no start symbol is specifically designated, the LHS of the first production is the start symbol.

2.2.2: Derivations

Watch how we can generate the string 7+4-5 beginning with the start symbol, applying productions, and stopping when no productions can be applied (because only terminals remain).

    list → list - digit
         → list - 5
         → list + digit - 5
         → list + 4 - 5
         → digit + 4 - 5
         → 7 + 4 - 5
  

This process of applying productions, starting with the start symbol and ending when only terminals are present is called a derivation and we say that the final string has been derived from the initial string (in this case the start symbol).

The set of all strings derivable from the start symbol is the language generated by the CFG

• It is important that you see that this context-free grammar generates precisely the set of infix expressions with single digits as operands (so 25 is not allowed) and + and - as operators.
• The way you get different final expressions is that you make different choices of which production to apply. There are 3 productions you can apply to list and 10 you can apply to digit.
• The result cannot have blanks since blank is not a terminal.
• The empty string is not possible since, starting from list, we cannot get to the empty string. If we wanted to include the empty string, we would add the production
  list → ε
• The idea is that the input language to the compiler is approximately the language generated by the grammar. It is approximate since I have ignored the scanner.

Given a grammar, parsing a string (of terminals) consists of determining if the string is in the language generated by the grammar. If it is in the language, parsing produces a derivation. If it is not, parsing reports an error.

The opposite of derivation is reduction. Given a production, the LHS produces or derives the RHS (a derivation) and the RHS is reduced to the LHS (a reduction).

Ignoring errors for the moment, parsing a string means reducing the string to the start symbol or equivalently deriving the string from the start symbol.

Homework: 1a, 1c, 2a-c (don't worry about justifying your answers).

Remark: Since we are in section 2.2 these questions are in section 2.2.7, the last subsection of section 2.2.

2.2.3: Parse trees

While deriving 7+4-5, one could produce the Parse Tree shown on the right.

You can read off the productions from the tree. For any internal (i.e., non-leaf) tree node, its children give the right hand side (RHS) of a production having the node itself as the LHS.

The leaves of the tree, read from left to right, is called the yield of the tree. We say that this string is derived from the (nonterminal at the) root, or is generated by the root, or can be reduced to the root. The tree on the right shows that 7+4-5 can be derived from list.

Homework: 1b

2.2.4: Ambiguity

An ambiguous grammar is one in which there are two or more parse trees yielding the same final string. We wish to avoid such grammars.

The grammar above is not ambiguous. For example 1+2+3 can be parsed only one way; the arithmetic must be done left to right. Note that I am not giving a rule of arithmetic, just of this grammar. If you reduced 2+3 to list you would be stuck since it is impossible to further reduce 1+list (said another way it is not possible to derive 1+list from the start symbol).

Remark:
The following is a wrong proof of ambiguity. Consider the grammar
S → A B
A → x
B → x
This grammar is ambiguous because we can derive the string x x in two ways
S → A B → A x → x x
S → A B → x B → x x
WRONG!!
There are indeed two derivations, but they have the same parse tree!
End of Remark.

Homework: 3 (applied only to parts a, b, and c of 2)

2.2.5: Associativity of operators

Our grammar gives left associativity. That is, if you traverse the parse tree in postorder and perform the indicated arithmetic you will evaluate the string left to right. Thus 8-8-8 would evaluate to -8. If you wished to generate right associativity (normally exponentiation is right associative, so 2**3**2 gives 512 not 64), you would change the first two productions to

  list → digit + list
  list → digit - list

Draw in class the parse tree for 7+4-5 with this new grammar.

2.2.6: Precedence of operators

We normally want * to have higher precedence than +. We do this by using an additional nonterminal to indicate the items that have been multiplied. The example below gives the four basic arithmetic operations their normal precedence unless overridden by parentheses. Redundant parentheses are permitted. Equal precedence operations are performed left to right.

  expr   → expr + term | expr - term | term
  term   → term * factor | term / factor | factor
  factor → digit | ( expr )
  digit  → 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9

Do the examples 1+2/3-4*5 and (1+2)/3-4*5 on the board.

Note how the precedence is enforced by the grammar; slick!

Start Lecture #2

Statements

Keywords are very helpful for distinguishing statements from one another.

    stmt → id := expr
    	 | if expr then stmt
    	 | if expr then stmt else stmt
    	 | while expr do stmt
    	 | begin opt-stmts end
    opt-stmts → stmt-list | ε
    stmt-list → stmt-list ; stmt | stmt
  

Remarks:

1. In the above example I underlined the nonterminals. This is not normally done. It is easy to tell the nonterminals; they are the symbols that appear on the LHS.
2. opt-stmts stands for optional statements. The begin-end block can be empty in some languages.
3. The ε (epsilon) stands for the empty string.
4. The use of epsilon productions will add complications.
5. Some languages do not permit empty blocks. For example, Ada has a null statement, which does nothing when executed, but avoids the need for empty blocks.
6. The above grammar is ambiguous!
7. The notorious dangling else problem.
8. How do you parse if x then if y then z=1 else z=2?

Homework: 4 a-d (for a the operands are digits and the operators are +, -, *, and /).

2.3: Syntax-Directed Translation

The idea is to specify the translation of a source language construct in terms of attributes of its syntactic components. The basic idea is use the productions to specify a (typically recursive) procedure for translation. For example, consider the production

    stmt-list → stmt-list ; stmt
  

1. Call ourselves recursively to process the right stmt-list (which is smaller). This will, say, generate code for all the statements in the right stmt-list.
2. Call the procedure for stmt, generating code for stmt.
3. Process the left stmt-list by combining the results for the first two steps as well as what is needed for the semicolon (a terminal, so we do not further delegate its actions). In this case we probably concatenate the code for the right stmt-list and stmt.

To avoid having to say the right stmt-list and the left stmt-list we write the production as

    stmt-list → stmt-list1 ; stmt
  

Question: Why won't this go on forever?
Answer: Eventually stmt-list₁ will consist of only one stmt and then the other production for stmt-list will be used.

2.3.1: Postfix Notation (An Example)

This notation is called postfix because the rule is operator after operand(s). Parentheses are not needed. The notation we normally use is called infix. If you start with an infix expression, the following algorithm will give you the equivalent postfix expression.

• Variables and constants are left alone.
• E op F becomes E' F' op, where E' and F' are the postfix of E and F respectively.
• ( E ) becomes E', where E' is the postfix of E.

One question is, given say 1+2-3, what are E, F and op? Does E=1+2, F=3, and op=-? Or does E=1, F=2-3 and op=+? This is the issue of precedence and associativity mentioned above. To simplify the present discussion we will start with fully parenthesized infix expressions.

Example: 1+2/3-4*5

1. Start with 1+2/3-4*5
2. Parenthesize (using standard precedence) to get (1+(2/3))-(4*5)
3. Apply the above rules to calculate P{(1+(2/3))-(4*5)}, where P{X} means convert the infix expression X to postfix.
   1. P{(1+(2/3))-(4*5)}
   2. P{(1+(2/3))} P{(4*5)} -
   3. P{1+(2/3)} P{4*5} -
   4. P{1} P{2/3} + P{4} P{5} * -
   5. 1 P{2} P{3} / + 4 5 * -
   6. 1 2 3 / + 4 5 * -

Example: Now do (1+2)/3-4*5

1. Parenthesize to get ((1+2)/3)-(4*5)
2. Calculate P{((1+2)/3)-(4*5)}
   1. P{((1+2)/3) P{(4*5)} -
   2. P{(1+2)/3} P{4*5) -
   3. P{(1+2)} P{3} / P{4} P{5} * -
   4. P{1+2} 3 / 4 5 * -
   5. P{1} P{2} + 3 / 4 5 * -
   6. 1 2 + 3 / 4 5 * -

2.3.2: Synthesized Attributes

We want to decorate the parse trees we construct with annotations that give the value of certain attributes of the corresponding node of the tree.

In November, we will use as input an algol/ada/C-like language and will have a .code attribute for many nodes with the property that .code contains the intermediate code that results from compiling the program corresponding to the leaves of the subtree rooted at this node. In particular,

However, it is now only September so our goal is more modest. We will do the example of translating infix to postfix using the same infix grammar as above. It uses standard arithmetic terminology where one multiplies and divides factors to obtain terms, which in turn are added and subtracted to form expressions.

  expr   → expr + term | expr - term | term
  term   → term * factor | term / factor | factor
  factor → digit | ( expr )
  digit  → 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9

This grammar supports parentheses, although our example 1+2/3-4*5 does not use them. On the right is a movie in which the parse tree is built from this example.

Question: Was this a top-down or bottom-up movie?

The attribute we will associate with the nodes is the postfix form of the string in the leaves below the node. In particular, the value of this attribute at the root is the postfix form of the entire source.

The book does a simpler grammar (no *, /, or parentheses) for a simpler example. You might find that one easier.

Syntax-Directed Definitions (SDDs)

Definition: A syntax-directed definition is a grammar together with semantic rules associated with the productions. These rules are used to compute attribute values. A parse tree augmented with the attribute values at each node is called an annotated parse tree.

For the bottom-up approach I will illustrate now, we annotate a node after having annotated its children. Thus the attribute values at a node can depend on the children of the node but not the parent of the node. We call these synthesized attributes, since they are formed by synthesizing the attributes of the children.

In chapter 5, when we study top-down annotations as well, we will introduce inherited attributes that are passed down from parents to children.

We specify how to synthesize attributes by giving the semantic rules together with the grammar. That is, we give the syntax directed definition.

We apply these rules bottom-up (starting with the geographically lowest productions, i.e., the lowest lines in the tree) and get the annotated graph shown on the right. The annotation are drawn in green.

Homework: Draw the annotated graph for (1+2)/3-4*5.

2.3.3: Simple Syntax-Directed Definitions

If the semantic rules of a syntax-directed definition all have the property that the new annotation for the left hand side (LHS) of the production is just the concatenation of the annotations for the nonterminals on the RHS in the same order as the nonterminals appear in the production, we call the syntax-directed definition simple. It is still called simple if new strings are interleaved with the original annotations. So the example just done is a simple syntax-directed definition.

Remark: SDD's feature semantic rules. We will soon learn about Translation Schemes, which feature a related concept called semantic actions. When one has a simple SDD, the corresponding translation scheme can be done without constructing the parse tree. That is, while doing the parse, when you get to the point where you would construct the node, you just do the actions. In the translation scheme corresponding to the present example, the action at a node is just to print the new strings at the appropriate points.

2.3.4: (depth-first) Tree Traversals

When performing a depth-first tree traversal, it is clear in what order the leaves are to be visited, namely left to right. In contrast there are several choices as to when to visit an interior (i.e. non-leaf) node. The traversal can visit an interior node

1. Before visiting any of its children.
2. Between visiting its children.
3. After visiting all of its children.

I do not like the book's pseudocode as I feel the names chosen confuse the traversal with visiting the nodes. I prefer the pseudocode below, which uses the following conventions.

• Comments are introduced by -- and terminate at the end of the line (as in the programming language Ada).
  
  
• Indenting is significant so begin/end or {} are not used (from the programming language family B2/ABC/Python)

    traverse (n : treeNode)
        if leaf(n)                      -- visit leaves once; base of recursion
           visit(n)
        else                            -- interior node, at least 1 child
           -- visit(n)                  -- visit node PRE visiting any children
           traverse(first child)        -- recursive call
           while (more children remain) -- excluding first child
               -- visit(n)              -- visit node IN-between visiting children
               traverse (next child)    -- recursive call
           -- visit(n)                  -- visit node POST visiting all children
  

Note the following properties

1. As written, with the last three visit()s commented out, only the leaves are visited and those visits are in left to right order.
   
   
2. If you uncomment just the first (interior node) visit, you get a preorder traversal, in which each node is visited before (i.e., pre) visiting any of its children.
   
   
3. If you uncomment just the last visit, you get a postorder traversal, in which each node is visited after (i.e., post) visiting all of its children.
   
   
4. If you uncomment only the middle visit, you get an inorder traversal, in which the node is visited (in-) between visiting its children.
   
   Inorder traversals are normally defined only for binary trees, i.e., trees in which every interior node has exactly two children. Although the code with only the middle visit uncommented works for any tree, we will, like everyone else, reserve the name inorder traversal for binary trees. In the case of binary search trees (everything in the left subtree is smaller than the root of that subtree, which in tern is smaller than everything in the corresponding right subtree) an inorder traversal visits the values of the nodes in (numerical) order.
   
   
5. If you uncomment two of the three visits, you get a traversal without a name.
   
   
6. If you uncomment all of the three visits, you get an Euler-tour traversal.

To explain the name Euler-tour traversal, recall that an Eulerian tour on a directed graph is one that traverses each edge once. If we view the tree on the right as undirected and replace each edge with two arcs, one in each direction, we see that the pink curve is indeed an Eulerian tour. It is easy to see that the curve visits the nodes in the order of the pseudocode (with all visits uncommented).

Normally, the Euler-tour traversal is defined only for a binary tree, but this time I will differ from convention and use the pseudocode above to define Euler-tour traversal for all trees.

Note the following points about our Euler-tour traversal.

• A node with k children is visited k+1 times. The diagram shows nodes with 0, 1, 2, and 3 children.
• In a binary tree, a leaf is visited once and an interior node is visited three times. This is one of the standard definitions of an Euler-tour traversal for a binary tree.
• The other standard definition has all nodes visited 3 times. For a leaf the three visits are in succession. Modifying the pseudocode to obtain this definition simply requires replacing the leaf visit with
  
  

  	visit(n); visit(n); visit(n)
        

Remarks

• Since, at this point in the course, we are considering only synthesized attributes, a postorder traversal will always yield a correct evaluation order for the attributes. This is so since synthesized attributes depend only on attributes of child nodes and a postorder traversal visits a node only after all the children have been visited (and hence all the child node attributes have been evaluated).
• In the general case (when not all attributes are synthesized), SDDs do not specify an evaluation order for the attributes of the parse tree. The requirement remains that each attribute is evaluated after all those that it depends on. This general case is quite difficult, and sometimes no such order is possible.

2.3.5: Translation schemes

The bottom-up annotation scheme just described generates the final result as the annotation of the root. In our infix to postfix example we get the result desired by printing the root annotation. Now we consider another technique that produces its results incrementally.

Instead of giving semantic rules for each production (and thereby generating annotations) we can embed program fragments called semantic actions within the productions themselves.

When drawn in diagrams (e.g., see the diagram below), the semantic action is connected to its node with a distinctive, often dotted, line. The placement of the actions determine the order they are performed. Specifically, one executes the actions in the order they are encountered in a depth-first traversal of the tree (the children of a node are visited in left to right order). Note that these action nodes are all leaves and hence they are encountered in the same order for both preorder and postorder traversals (and inorder and Euler-tree order).

Definition: A syntax-directed translation scheme is a context-free grammar with embedded semantic actions.

In the SDD for our infix to postfix translator, the parent either

1. takes the attribute of its only child or
2. concatenates the attributes left to right of its several children and adds something at the end.

Emitting a Translation

The table on the right gives the semantic actions corresponding to a few of the rows of the table above. Note that the actions are enclosed in {}. The corresponding semantic rules are given as well.

It is redundant to give both semantic actions and semantic rules; in practice, we use one or the other. In this course we will emphasize semantic rules, i.e. syntax directed definitions (SDDs). I show both the rules and the actions in a few tables just so that we can see the correspondence.

The diagram for 1+2/3-4*5 with attached semantic actions is shown on the right.

Given an input, e.g. our favorite 1+2/3-4*5, we just do a (left-to-right) depth first traversal of the corresponding diagram and perform the semantic actions as they occur. When these actions are print statements as above, we are said to be emitting the translation.

Since the actions are all leaves of the tree, they occur in the same order for any depth-first (left-to-right) traversal (e.g., postorder, preorder, or Euler-tour order).

Do on the board a depth first traversal of the diagram, performing the semantic actions as they occur, and confirm that the translation emitted is in fact 123/+45*-, the postfix version of 1+2/3-4*5

Homework: Produce the corresponding diagram for (1+2)/3-4*5.

Prefix to infix translation

When we produced postfix, all the prints came at the end (so that the children were already printed). The { action }'s do not need to come at the end. We illustrate this by producing infix arithmetic (ordinary) notation from a prefix source.

In prefix notation the operator comes first. For example, +1-23 evaluates to zero and +-123 evaluates to 2. Consider the following grammar, which generates the simple language of prefix expressions consisting of addition and subtraction of digits between 1 and 3 without parentheses (prefix notation and postfix notation do not use parentheses).

    P → + P P | - P P | 1 | 2 | 3
  

The resulting parse tree for +1-23 with the semantic actions attached is shown on the right. Note that the output language (infix notation) has parentheses.

The table below shows both the semantic actions and rules used by the translator. As mentioned previously, one normally does not use both actions and rules.

First do a preorder traversal of the tree and see that you get 1+(2-3). In fact you don't get that answer, but instead get a fully parenthesized version that is equivalent.

Next start a postorder traversal and see that it produces the same output (i.e., executes the same prints in the same order).

Question: What about an Euler-tour order?

Answer: The same result. For all traversals, the leaves are printed in left to right order and all the semantic actions are leaves.

Finally, pretend the prints aren't there, i.e., consider the unannotated parse tree and perform a postorder traversal, evaluating the semantic rules at each node encountered. Postorder is needed (and sufficient) since we have synthesized attributes and hence having child attributes evaluated prior to evaluating parent attributes is both necessary and sufficient to ensure that whenever an attribute is evaluated all the component attributes have already been evaluated. (It will not be so easy in chapter 5, when we have inherited attributes as well.)

Homework: 2.

2.4: Parsing

Objective: Given a string of tokens and a grammar, produce a parse tree yielding that string (or at least determine if such a tree exists).

We will learn both top-down (begin with the start symbol, i.e. the root of the tree) and bottom up (begin with the leaves) techniques.

In the remainder of this chapter we just do top down, which is easier to implement by hand, but is less general. Chapter 4 covers both approaches.

Tools (so called parser generators) often use bottom-up techniques.

In this section we assume that the lexical analyzer has already scanned the source input and converted it into a sequence of tokens.

2.4.1: Top-down parsing

Consider the following simple language, which derives a subset of the types found in the (now somewhat dated) programming language Pascal. I do not assume you know pascal.

We have two nonterminals, type, which is the start symbol, and simple, which represents the simple types.

There are 8 terminals, which are tokens produced by the lexer and correspond closely with constructs in pascal itself. Specifically, we have.

1. integer and char
2. id for identifier
3. array and of used in array declarations
4. ↑ meaning pointer to
5. num for a (positive whole) number
6. dotdot for .. (used to give a range like 6..9)

The productions are

    type   → simple
    type   → ↑ id
    type   → array [ simple ] of type
    simple → integer
    simple → char
    simple → num dotdot num
  

Parsing is easy in principle and for certain grammars (e.g., the one above) it actually is easy. We start at the root since this is top-down parsing and apply the two fundamental steps.

1. At the current (nonterminal) node, select a production whose LHS is this nonterminal and whose RHS matches the input at this point. Make the RHS the children of this node (one child per RHS symbol).
2. Go to the next node needing a subtree.

When programmed this becomes a procedure for each nonterminal that chooses a production for the node and calls procedures for each nonterminal in the RHS of that production. Thus it is recursive in nature and descends the parse tree. We call these parsers recursive descent.

The big problem is what to do if the current node is the LHS of more than one production. The small problem is what do we mean by the next node needing a subtree.

The movie on the right, which succeeds in parsing, works by tossing 2 ounces of pixie dust into the air and choosing the production onto which the most dust falls. (An alternative interpretation is given below.)

The easiest solution to the big problem would be to assume that there is only one production having a given terminal as LHS. There are two possibilities

1. No circularity. For example

   	expr → term + term - 9
   	term → factor / factor
   	factor → digit
   	digit → 7
         

   But this is very boring. The only possible sentence is 7/7+7/7-9
   
   
2. Circularity

   	expr → term + term
   	term → factor / factor
   	factor → ( expr )
         

   This is even worse; there are no (finite) sentences. Only an infinite sentence beginning (((((((((.

So this won't work. We need to have multiple productions with the same LHS.

How about trying them all? We could do this! If we get stuck where the current tree cannot match the input we are trying to parse, we would backtrack.

Instead, we will look ahead one token in the input and only choose productions that can yield a result starting with this token. Furthermore, we will (in this section) restrict ourselves to predictive parsing in which there is only one production that can yield a result starting with a given token. This solution to the big problem also solves the small problem. Since we are trying to match the next token in the input, we must choose the leftmost (nonterminal) node to give children to.

2.4.2: Predictive parsing

Let's return to pascal array type grammar and consider the three productions having type as LHS. Remember that, even when I write the short form

    type → simple | ↑ id | array [ simple ] of type
  

For each production P we wish to construct the set FIRST(P) consisting of those tokens (i.e., terminals) that can appear as the first symbol of a string derived from the RHS of P.

FIRST is actually defined on strings not productions. When I write FIRST(P), I really mean FIRST(RHS). Similarly, I often say the first set of the production P when I should really say the first set of the RHS of the production P. Formally, we proceed as follows.

Let α be a string of terminals and/or nonterminals. FIRST(α) is the set of terminals that can appear as the first symbol in a string of terminals derived from α. If α is ε or α can derive ε, then ε is in FIRST(α)

So given α we find all strings of terminals that can be derived from α and pick off the first terminal.

Question: How do we calculate FIRST(α)?

Answer: Wait until chapter 4 for a formal algorithm. For these simple examples it is reasonably clear.

Definition: Let r be the RHS of a production P. FIRST(P) is FIRST(r).

To use predictive parsing, we make the following

Assumption: Let P and Q be two productions with the same LHS, Then FIRST(P) and FIRST(Q) are disjoint. Thus, if we know both the LHS and the token that must be first, there is (at most) one production we can apply. BINGO!

An example of predictive parsing

This table gives the FIRST sets for our pascal array type example.

Make sure that you understand how this table was derived. It is not yet clear how to calculate FIRST for a complicated example. We will learn the general procedure in chapter 4.

Note that the three productions with type as LHS have disjoint FIRST sets. Similarly the three productions with simple as LHS have disjoint FIRST sets. Thus predictive parsing can be used. We process the input left to right and call the current token lookahead since it is how far we are looking ahead in the input to determine the production to use. The movie on the right shows the process in action.

Homework:

A. Construct the FIRST sets for

    rest → + term rest | - term rest | term
    term → 1 | 2 | 3
  

End of Homework.

2.4.3: When to Use ε-productions

Not all grammars are as friendly as the last example. The first complication is when ε occurs as a RHS. If this happens or if the RHS can generate ε, then ε is included in FIRST.

But ε would always match the current input position!

The rule is that if lookahead is not in FIRST of any production with the desired LHS, we use the (unique!) production (with that LHS) that has ε in FIRST.

The text does a C instead of a pascal example. The productions are

    stmt → expr ;
    	 | if ( expr ) stmt
    	 | for ( optexpr ; optexpr ; optexpr ) stmt
    	 | other
    optexpr → expr | ε
  

For completeness, on the right is the beginning of a movie for the C example. Note the use of the ε-production at the end since no other entry in FIRST will match ;

Once again, the full story will be revealed in chapter 4 when we do parsing in a more complete manner.

2.4.4: Designing a Predictive Parser

Predictive parsers are fairly easy to construct as we will now see. Since they are recursive descent parsers we go top-down with one procedure for each nonterminal. Do remember that to use predictive parsing, we must have disjoint FIRST sets for all the productions having a given nonterminal as LHS.

1. For each nonterminal, write a procedure that chooses the unique(!) production having lookahead in its FIRST set. Use the ε production if no other production matches. If no production matches and there is no ε production, the parse fails.
2. Having chosen a production, these procedures then mimic the RHS of the production. They call procedures for each nonterminal and call match for each terminal.
3. Write a procedure match(terminal) that advances lookahead to the next input token after confirming that the previous value of lookahead equals the terminal argument.
4. Write a main program that initializes lookahead to the first input token and invokes the procedure for the start symbol.

The book has code at this point, which you should read.

2.4.5: Left Recursion

Another complication. Consider

    expr → expr + term
    expr → term
  

For the first production the RHS begins with the LHS. This is called left recursion. If a recursive descent parser would pick this production, the result would be that the next node to consider is again expr and the lookahead has not changed. An infinite loop occurs. (Also note that the first sets are not disjoint.)

Note that this is NOT a problem with the grammar per se, but is a limitation of predictive parsing. For example if we had the additional production

    term → x
  

    x + x
  

If the grammar were instead

    expr → term + expr
    expr → term
  

Consider, instead of the original (left-recursive) grammar, the following replacement

    expr → term rest
    rest → + term rest
    rest → ε
  

Both sets of productions generate the same possible token strings, namely

    term + term + ... + term
  

Using this technique to eliminate left-recursion will (next month) make it harder for us to get left associativity for arithmetic, but we shall succeed!

In general, for any nonterminal A, and any strings α, and β (α and β cannot start with A), we can replace the pair of productions

    A → A α | β
  

    A → β R
    R → α R | ε
  

For the example above A is expr, R is rest, α is + term, and β is term.

Yes, this looks like magic.
Yes, there are more general possibilities.
We will have more to say in chapter 4.

2.5: A Translator for Simple Expressions

Objective: An infix to postfix translator for expressions. We start with just plus and minus, specifically the expressions generated by the following grammar. We include a set of semantic actions with the grammar. Note that finding a grammar for the desired language is one problem, constructing a translator for the language, given a grammar, is another problem. We are tackling the second problem.

    expr → expr + term { print('+') }
    expr → expr - term { print('-') }
    expr → term
    term → 0           { print('0') }
    . . .
    term → 9           { print('9') }
  

One problem we must solve is that this grammar is left recursive.

2.5.1: Abstract and Concrete Syntax

Often one prefers not to have superfluous nonterminals as they make the parsing less efficient. That is why we don't say that a term produces a digit and a digit produces each of 0,...,9. Ideally the syntax tree would just have the operators + and - and the 10 digits 0,1,...,9. That would be called the abstract syntax tree. A parse tree coming from a grammar is technically called a concrete syntax tree.

2.5.2: Adapting the Translation Scheme

We eliminate the left recursion as we did in 2.4. This time there are two operators + and - so we replace the triple

    A → A α | A β | γ
  

    A → γ R
    R → α R | β R | ε
  

This time we have actions so, for example

    α is + term { print('+') }
  

    expr → term rest
    rest → + term { print('+') } rest
         | - term { print('-') } rest
         | ε
    term → 0           { print('0') }
    . . .
         | 9           { print('9') }
  

2.5.3: Procedures for the Nonterminals expr, term, and rest

The C code is in the book. Note the else ; in rest(). This corresponds to the epsilon production. As mentioned previously. The epsilon production is only used when all others fail (that is why it is the else arm and not the then or the else if arms).

2.5.4: Simplifying the translator

These are (useful) programming techniques.

The complete program

The program in Java is in the book.

Start Lecture #3

Remark: Summary of end of last time.

1. We have a grammar for the simple expressions. It has no ε-productions (good news), but is left recursive (bad news). First eliminate left recursion and then use predictive parsing to write a program (a parser) that constructs a parse tree for any input string (i.e., for any infix expression).
2. But we can do better. We gave an SDT (i.e. gave actions). Again you can eliminate the left recursion (done in the notes). Now, when your parser constructs the parse tree it is more, it has the print statements as additional leaves.
3. Then you just do a lab 1 traversal (post/pre/EulerTour-order) on this enhanced tree with visit(regularNode) is a nop and visit(printNode) just does the print.
4. The result is the postfix for the given infix.
5. Thus, you have constructed a 2 phase compiler (enhanced parser; tree walker) from infix to postfix.
6. In fact for this simple example you can further modify the parser to not actually produce the parse tree (saving memory). However, we won't do this. Compilers for real programming language can't do this spacing saving since the grammars/SDDs/SDTs are not as simple as this example.

2.6: Lexical Analysis

The purpose of lexical analysis is to convert a sequence of characters (the source) into a sequence of tokens. A lexeme is the sequence of characters comprising a single token.

Note that (following the book) we are going out of order. In reality, the lexer operates on the input and the resulting token sequence is the input to the parser. The reason we were able to produce the translator in the previous section without a lexer is that all the tokens were just one character (that is why we had just single digits).

Actually, you never need a lexer. Anything that a lexer can do a parser can do. But lexers are smaller and for software engineering and other reasons are normally used.

2.6.1: Removal of White space and comments

These do not become tokens so that the parser need not worry about them.

2.6.2: Reading ahead

Consider distinguishing x<y from x<=y.

After reading the < we must read another character. If it is y, we have found our token (<). However, we must unread the y so that when asked for the next token, we will start at y. If it is never more than one extra character that must be examined, a single char variable would suffice. A more general solution is discussed in the next chapter (Lexical Analysis).

2.6.3: Constants

This chapter considers only numerical integer constants. They are computed one digit at a time using the formula

    value:=10*value+digit.
  

The value of the constant can be considered the attribute of the token named num. Alternatively, the attribute can be a pointer/index into the symbol table entry for the number (or into a numbers table).

2.6.4: Recognizing identifiers and keywords

The C statement
    sum = sum + x;
contains 6 tokens. The scanner (aka lexer; aka lexical analyzer) will convert the input into
    id = id + id ;
(id standing for identifier).
Although there are three id tokens, the first and second represent the lexeme sum; the third represents x. These two different lexemes must be distinguished.

A related distinction occurs with language keywords, for example then, which are syntactically the same as identifiers. The symbol table is used to accomplishes both distinctions. We assume (as do most modern languages) that the keywords are reserved, i.e., cannot be used as program variables. The we simply initialize the symbol table to contain all these reserved words and mark them as keywords. When the lexer encounters a would-be identifier and searches the symbol table, it finds out that the string is actually a keyword.

As mentioned previously care must be taken when one lexeme is a proper subset of another. Consider
    x<y versus x<=y
When the < is read, the scanner needs to read another character to see if it is an =. But if that second character is y, the current token is < and the y must be pushed back onto the input stream so that the configuration is the same after scanning < as it is after scanning <=.

Also consider then versus thenewvalue, one is a keyword and the other an id.

2.6.5: A lexical analyzer

A Java program is given. The book, but not the course, assumes knowledge of Java.

Since the scanner converts digits into num's we can shorten the grammar above. Here is the shortened version before the elimination of left recursion. Note that the value attribute of a num is its numerical value.

    expr   → expr + term    { print('+') }
    expr   → expr - term    { print('-') }
    expr   → term
    term   → num            { print(num.value) }
  

    expr   → expr + term    { print('+') }
    expr   → expr - term    { print('-') }
    expr   → term
    term   → factor
    factor → ( expr ) | num { print(num,value) }
  

The factor() procedure follows the familiar recursive descent pattern: Find a production with factor as LHS and lookahead in FIRST, then do what the RHS says. That is, call the procedures corresponding to the nonterminals, match the terminals, and execute the semantic actions.

Note that we are now able to consider constants of more than one digit.

2.7: Incorporating a symbol table

The symbol table is an important data structure for the entire compiler. One example of its use is that the semantic actions or rules associated with declarations set the type field of the symbol table entry. Subsequent semantic actions or rules associated with expression evaluation use this type information. For the simple infix to postfix translator (which is typeless), the table is primarily used to store and retrieve <lexeme,token> pairs.

2.7.1: Symbol Table per Scope

There is a serious issue here involving scope. We will learn that lexers are based on regular expressions; whereas parsers are based on the stronger, but more expensive, context-free grammars. Regular expressions are not powerful enough to handle nested scopes. So, if the language you are compiling supports nested scopes, the lexer can only construct the <lexeme,token> pairs. The parser converts these pairs into a true symbol table that reflects the nested scopes. If the language is flat, the scanner can produce the symbol table.

The idea for a language with nested scopes is that, when entering a block, a new symbol table is created. Each such table points to the one immediately outer. This structure supports the most-closely nested rule for symbols: a symbol is in the scope of most-closely nested declaration. This gives rise to a tree of tables.

Interface

• Create table: A new table is created and points to the immediately outer table, which is passed as a argument.
• Insert entry (in the current table).
• Retrieve entry (from the most-closely nested table in which it appears).

Reserved keywords

Simply insert them into the symbol table prior to examining any input. Then they can be found when used correctly and, since their corresponding token will not be id, any use of them where an identifier is required can be flagged. For example a lexer for a C-like language would have insert(int) performed prior to scanning the input.

2.7.2: The Use of Symbol Tables

Below is the grammar for a stripped down example showing nested scopes. The language consists just of nested blocks, a weird mixture of C- and ada-style declarations (specifically, type colon identifier), and trivial statements consisting of just an identifier.

    program → block
    block   → { decls stmts }     -- { } are terminals not actions
    decls   →  decls decl | ε     -- study this one
    decl    → type : id ;
    stmts   → stmts stmt | ε      -- same idea, a list
    stmt    → block | factor ;    -- enables nested blocks
    factor  → id
  

One possible program in this language is

    { int : x ;  float : y ;
      x ; y ;
      { float : x ;
        x ; y ;
      }
      { int : y ;
        x ; y;
      }
      x ; y ;
    }
  

To show that we have correctly parsed the input and obtained its meaning (i.e., performed semantic analysis), we present a translation scheme that digests the declarations and translates the statements so that the above example becomes

{ int; float; { float; float; } { int; int; } }
  

The translation scheme, slightly modified from the book page 90, is shown on the right. First a formatting comment.

This translation scheme looks weird, but is actually a good idea (of the authors): it reconciles the two goals of respecting the ordering and nonetheless having the actions all in one column.

Recall that the placement of the actions within the RHS of the production is significant. The parse tree with its actions is processed in a depth first manner so that the actions are performed in left to right order. Thus an action is executed after all the subtrees rooted by parts of the RHS to the left of the action and is executed before all the subtrees rooted by parts of the RHS to the right of the action.

Consider the first production. We want the action to be executed before processing block. Thus the action must precede block in the RHS. But we want the actions in the right column. So we split the RHS over several lines and place an action in the rightmost column of the line that puts in the right order.

The second production has some semantic actions to be performed at the start of the block, and others to be performed at the bottom.

To fully understand the details, you must read the book; but we can see how it works. A new Env initializes a new symbol table; top.put inserts into the symbol table of the current environment top; top.get retrieves from that symbol table.

In some sense the star of the show is the simple production
    factor → id
together with its semantic actions. These actions look up the identifier in the (correct!) symbol table and print out the type name.

Question: Why do we have the trivial-looking production

    program → block
  

Hard Question: I prefer ada-style declarations, which are of the form
    identifier : type
What problem would have occurred had I done so here and how does one solve that problem?
Answer: Both a stmt and a decl would start with id and thus the FIRST sets would not be disjoint. The result would be that need more than one token lookahead to see when the decls end and the stmts begin. The fix is to left-factor the grammar as we will learn in chapter 4.

2.8: Intermediate Code Generation

2.8.1: Two kinds of Intermediate Representations

There are two important forms of intermediate representations.

• Trees, especially parse trees and syntax trees.
• Linear, especially three-address code

Another (but less common) name for parse tree is concrete syntax tree. Similarly another (also less common) name for syntax tree is abstract syntax tree.

Very roughly speaking, (abstract) syntax trees are parse trees reduced to their essential components, and three address code looks like assembler without the concept of registers.

2.8.2: Construction of (Abstract) Syntax Trees

Remarks:

1. Despite the words below, your future lab assignments will likely not require producing abstract syntax trees. Instead, you will be producing concrete syntax trees (parse trees). I may include an extra-credit part of some labs that will ask for abstract syntax trees.
2. Note however that real compilers do not produce parse trees since such trees are larger and have no extra information that the compiler needs. If the compiler produces a tree (many do), it produces an abstract syntax tree.
3. The reason I will not require your labs to produce the smaller trees is that to do so it is helpful to understand semantic rules and semantic actions, which come later in the course. Of course, authors of real compilers have already completed the course before starting the design so this consideration does not apply to them. :-)

Consider the production

    while-stmt → while ( expr ) stmt ;
  

To generate this while node, we execute

    new While(x,y)
  

Syntax Trees for Statements

The book has an SDD on page 94 for several statements. The part for while reads
  stmt → while ( expr ) stmt₁   { stmt.n = new While(expr.n, stmt₁.n); }
The n attribute gives the syntax tree node.

Representing Blocks in Syntax Trees

Fairly easy

    stmt  → block          { stmt.n = block.n }
    block → { stmts }      { block.n = stmts.n }
  

    while ( x == 5 ) {
       blah
       blah
       more
    }
  

1. The tree for x==5.
2. The tree for blah blah more.

Syntax trees for Expressions

When parsing we need to distinguish between + and * to insure that 3+4*5 is parsed correctly, reflecting the higher precedence of *. However, once parsed, the precedence is reflected in the tree itself (the node for + has the node for * as a child). The rest of the compiler treats + and * largely the same so it is common to use the same node label, say OP, for both of them. So we see
    term → term₁ * factor     { term.n = new Op('*', term₁.n, factor.n); }

Note, however, that the SDD (Figure 2.39) essentially constructs both the parse tree and the syntax tree. That latter is constructed as the attributes in the former.

2.8.3: Static Checking

Static checking refers to checks performed during compilation; whereas, dynamic checking refers to those performed at run time. Examples of static checks include

• Syntactic checks such as avoiding multiple declarations of the same identifier in the same scope.
• Type checks.

      struct symtableType {
        char lexeme[BIGNUMBER];
        int  token;
      } symtable[ANOTHERBIGNUMBER];
    

L-values and R-values

Consider Q := Z; or A[f(x)+B*D] := g(B+C*h(x,y));. I am using [] for array reference and () for function call).

From a macroscopic view, executing either of these assignments has three components.

1. Evaluate the left hand side (LHS) to obtain an l-value.
2. Evaluate the RHS to obtain an r-value.
3. Perform the assignment.

Note the differences between L-values, quantities that can appear on the LHS of an assignment, and and R-values, quantities that can appear only on the RHS.

• An l-value corresponds to an address or a location.
• An r-value corresponds to a value.
• Neither 12 nor s+t can be used as an l-value, but both are legal r-values.

Static checking is used to insure that R-values do not appear on the LHS.

Type Checking

These checks assure that the type of the operands are expected by the operator. In addition to flagging errors, this activity includes

• Coercions. The automatic conversion of one type to another. Later in this course we will employ the function widen(a,t,w) in certain semantic rules. Widen(x,int,double) would for example generate the intermediate code needed to convert x of type int into a quantity of type double.
• Overloading. In Java, Ada, and other languages, the same symbol can have different meanings depending on the types of the operands. Static checks are used to determine the correct operation, or signal an error if none exists

2.8.4: Three-Address Code

These are primitive instructions that have one operator and (up to) three operands, all of which are addresses. One address is the destination, which receives the result of the operation; the other two addresses are the sources of the values to be operated on.

Perhaps the clearest way to illustrate the (up to) three address nature of the instructions is to write them as quadruples or quads.

    ADD        x y z
    MULT       a b c
    ARRAY_L    q r s
    ARRAY_R    e f g
    ifTrueGoto x L
    COPY       r s
  

    x = y + z
    a = b * c
    q[r] = s
    e = f[g]
    ifTrue x goto L
    r = s
  

Translating Statements

We do this and the next section much slower and in much more detail later in the course.

Here is the if example from the book, which is somewhat Java intensive.

    class If extends Stmt {
       Expr E; Stmt S;
       public If(Expr x, Stmt y) { E = x;  S = y;  after = newlabel(); }
       public void gen() {
          Expr n = E.rvalue();
          emit ("ifFalse " + n.toString() + " goto " + after);
          S.gen();
          emit(after + ":");
       }
    }
  

The idea is that we are to translate the statement

    if expr then stmt
  

1. A block of code to compute the expression placing the result into x.
2. The single line   ifFalse x goto after.
3. A block of code for the statement(s) inside the then
4. The label after.

The constructor for an IF node is called with nodes for the expression and statement. These are saved.

When the entire tree is constructed, the main code calls gen() of the root. As we see for IF, gen of a node invokes gen of the children.

Translating Expressions

I am just illustrating the simplest case

    Expr rvalue(x : Expr) {
       if (x is an Id or Constant node) return x;
       else if (x is an Op(op, y, z) node) {
	  t = new temporary;
	  emit string for t = rvalue(y) op rvalue(z);
	  return a new node for t;
       else read book for other cases
  

Better Code for Expressions

So called optimization (the result is far from optimal) is a huge subject that we barely touch. Here are a few very simple examples. We will cover these since they are local optimizations, that is they occur within a single basic block (a sequence of statements that execute without any jumps).

1. For a Java assignment statement x = x + 1; we would generate two three-address instructions

   	temp = x + 1
   	x    = temp
         

   The can be combined into the three-address instruction x = x + 1, providing there are no further uses of the temporary.
2. Common subexpressions occurring in two different expressions, need be computed only once.

      lvalue y
      push 7
      rvalue xx
      *
      push 6
      rvalue z
      rvalue w
      +
      *
      +
      :=
    

      stmt → id := expr
        { stmt.t := 'lvalue' || id.lexime || expr.t || := }
    

      stmt → if expr then stmt1 { out := newlabel();
                               stmt.t := expr.t || 'gofalse' out || stmt1.t || 'label' out
    

      stmt → if
	 expr      { out := newlabel; emit('gofalse', out); }
	 then
         stmt1     { emit('label', out) }
    

      procedure stmt
        integer test, out;
        if lookahead = id then       // first set is {id} for assignment
          emit('lvalue', tokenval);  // pushes lvalue of lhs
          match(id);                 // move past the lhs]
          match(':=');               // move past the :=
          expr;                      // pushes rvalue of rhs on tos
          emit(':=');                // do the assignment (Omitted in book)
        else if lookahead = 'if' then
          match('if');               // move past the if
          expr;                      // pushes boolean on tos
          out := newlabel();
          emit('gofalse', out);      // out is integer, emit makes a legal label
          match('then');             // move past the then
          stmt;                      // recursive call
          emit('label', out)         // emit again makes out legal
        else if ...                  // while, repeat/do, etc
        else error();
      end stmt;
    

       start → list eof
        list → expr ; list
        list →  ε                   // would normally use | as below
        expr → expr + term      { print('+') }
             | expr - term      { print('-'); }
             | term
        term → term * factor    { print('*') }
             | term / factor    { print('/') }
             | term div factor  { print('DIV') }
             | term mod factor  { print('MOD') }
             | factor
      factor → ( expr )
             | id               { print(id.lexeme) }
             | num              { print(num.value) }
    

            start → list eof
    	 list → expr ; list
    	      | ε
    	 expr → term moreterms
        moreterms → + term { print('+') } moreterms
    	      | - term { print('-') } moreterms
    	      | ε
             term | factor morefactors
      morefactors → * factor { print('*') } morefactors
    	      | / factor { print('/') } morefactors
    	      | div factor { print('DIV') } morefactors
    	      | mod factor { print('MOD') } morefactors
    	      | ε
           factor → ( expr )
                  | id               { print(id.lexeme) }
                  | num              { print(num.value) }
    

    term() {
      int t;
      factor();
      // now we should call morefactorsl(), but instead code it inline
      while(true)              // morefactor nonterminal is right recursive
         switch (lookahead) {  // lookahead set by match()
         case '*': case '/': case DIV: case MOD: // all the same
            t = lookahead;     // needed for emit() below
            match(lookahead)   // skip over the operator
            factor();          // see grammar for morefactors
            emit(t,NONE);
            continue;          // C semantics for case
         default:              // the epsilon production
            return;
  

Two Questions

1. How come this compiler was so easy?
2. Why isn't the final exam next week?

One reason is that much was deliberately simplified. Specifically note that

• No real machine code generated (no back end).
• No optimizations (improvement to generated code).
• FIRST sets disjoint.
• No semantic analysis.
• Input language very simple.
• Output language very simple and closely related to input.

Also, I presented the material way too fast to expect full understanding.

Chapter 3: Lexical Analysis

Homework: Read chapter 3.

Two methods to construct a scanner (lexical analyzer).

1. By hand, beginning with a diagram of what lexemes look like. Then write code to follow the diagram and return the corresponding token and possibly other information.
2. Feed the patterns describing the lexemes to a lexer-generator, which then produces the scanner. The historical lexer-generator is Lex; a more modern one is flex.

Note that the speed (of the lexer not of the code generated by the compiler) and error reporting/correction are typically much better for a handwritten lexer. As a result most production-level compiler projects write their own lexers.

3.1: The Role of the Lexical Analyzer

The lexer is called by the parser when the latter is ready to process another token.

The lexer also might do some housekeeping such as eliminating whitespace and comments. Some call these tasks scanning, but others user the term scanner for the entire lexical analyzer.

After the lexer, individual characters are no longer examined by the compiler; instead tokens (the output of the lexer) are used.

3.1.1: Lexical Analysis Versus Parsing

Why separate lexical analysis from parsing? The reasons are basically software engineering concerns.

1. Simplicity of design. When one detects a well defined subtask (produce the next token), it is often good to separate out the task (modularity).
2. Efficiency. With the task separated out, it is easier to apply specialized techniques.
3. Portability. Only the lexer need communicate with the outside.

3.1.2: Tokens, Patterns, and Lexemes

• A token is a <name,attribute> pair. These are what the parser processes. The attribute might actually be a tuple of several attributes or a pointer to a table entry, which itself might contain many components.
• A pattern describes the character strings for the lexemes of the token. For example a letter followed by a (possibly empty) sequence of letters and digits.
• A lexeme for a token is a sequence of characters that matches the pattern for the token.

Note the circularity of the definitions for lexeme and pattern.

Common token classes.

1. One for each keyword. The pattern is trivial.
2. One for each operator or class of operators. A typical class is the comparison operators. Note that these have the same precedence. We might have + and - as the same token, but not + and *.
3. One for all identifiers (e.g. variables, user defined type names, etc).
4. Constants (i.e., manifest constants) such as 6 or hello, but not a constant identifier such as quantum in the Java statement.
   static final int quantum = 3;. There might be one token for integer constants, a different one for real, another for string, etc.
5. One for each punctuation symbol.

3.1.3: Attributes for Tokens

We saw an example of attributes in the last chapter.

For tokens corresponding to keywords, attributes are not needed since the name of the token tells everything. But consider the token corresponding to integer constants. Just knowing that the we have a constant is not enough, subsequent stages of the compiler need to know the value of the constant. Similarly for the token identifier we need to distinguish one identifier from another. The normal method is for the attribute to specify the symbol table entry for this identifier.

We really shouldn't say symbol table. As mentioned above if the language has scoping (nested blocks) the lexer can't construct the symbol table, but just makes a table of <lexeme,token> pairs, which the parser later converts into a proper symbol table (or tree of tables).

Homework: 1.

3.1.4: Lexical Errors

We saw in this movie an example where parsing got stuck because we reduced the wrong part of the input string. We also learned about FIRST sets that enabled us to determine which production to apply when we are operating left to right on the input. For predictive parsers the FIRST sets for a given nonterminal are disjoint and so we know which production to apply. In general the FIRST sets might not be disjoint so we have to try all the productions whose FIRST set contains the lookahead symbol.

All the above assumed that the input was error free, i.e. that the source was a sentence in the language. What should we do when the input is erroneous and we get to a point where no production can be applied?

In many cases this is up to the parser to detect/repair. Sometimes, however, the lexer is stuck because there are no patterns that match the input at this point.

The simplest solution is to abort the compilation stating that the program is wrong, perhaps giving the line number and location where the lexer and/or parser could not proceed.

We would like to do better and at least find other errors. We could perhaps skip input up to a point where we can begin anew (e.g. after a statement ending semicolon), or perhaps make a small change to the input around lookahead so that we can proceed.

3.2: Input Buffering

Determining the next lexeme often requires reading the input beyond the end of that lexeme. For example, to determine the end of an identifier normally requires reading the first whitespace character after it. Also just reading > does not determine the lexeme as it could also be >=. When you determine the current lexeme, the characters you read beyond it may need to be read again to determine the next lexeme.

3.3: Specification of Tokens

The chapter turns formal and, in some sense, the course begins. The book is fairly careful about finite vs infinite sets and also uses (without a definition!) the notion of a countable set. (A countable set is either a finite set or one whose elements can be put into one to one correspondence with the positive integers. That is, it is a set whose elements can be counted. The set of rational numbers, i.e., fractions in lowest terms, is countable; the set of real numbers is uncountable, because it is strictly bigger, i.e., it cannot be counted.) We should be careful to distinguish the empty set φ from the empty string ε. Formal language theory is a beautiful subject, but I shall suppress my urge to do it right and try to go easy on the formalism.

3.3.1: Strings and Languages

We will need a bunch of definitions.

Definition: An alphabet is a finite set of symbols.

Example: {0,1}, presumably φ (uninteresting), {a,b,c} (typical for exam questions), ascii, unicode, latin-1.

Definition: A string over an alphabet is a finite sequence of symbols from that alphabet. Strings are often called words or sentences.

Example: Strings over {0,1}: ε, 0, 1, 111010. Strings over ascii: ε, sysy, the string consisting of 3 blanks.

Definition: The length of a string is the number of symbols (counting duplicates) in the string.

Example: The length of allan, written |allan|, is 5.

Definition: A language over an alphabet is a countable set of strings over the alphabet.

Example: All grammatical English sentences with five, eight, or twelve words is a language over ascii.

Definition: The concatenation of strings s and t is the string formed by appending the string t to s. It is written st.

Example: εs = sε = s for any string s.

We view concatenation as a product (see Monoid in wikipedia http://en.wikipedia.org/wiki/Monoid). It is thus natural to define s⁰=ε and sⁱ⁺¹=sⁱs.

Example: s¹=s, s⁴=ssss.

More string terminology

A prefix of a string is a portion starting from the beginning and a suffix is a portion ending at the end. More formally,

Definitions:

1. A prefix of s is any string obtained from s by removing (possibly zero) characters from the end of s.
2. A suffix is defined analogously.
3. A substring of s is obtained by deleting a prefix and a suffix.

Note: Any prefix or suffix is a substring.

Examples: If s is 123abc, then

1. s itself and ε are each a prefix, suffix, and a substring.
2. 12 are 123a are prefixes and substrings.
3. 3abc is a suffix and a substring.
4. 23a is a substring

Definitions: A proper prefix of s is a prefix of s other than ε and s itself. Similarly, proper suffixes and proper substrings of s do not include ε and s.

Definition: A subsequence of s is formed by deleting (possibly zero) positions from s. We say positions rather than characters since s may for example contain 5 occurrences of the character Q and we only want to delete a certain 3 of them.

Example: issssii is a subsequence of Mississippi.

Note: Any substring is a subsequence.

Homework: 3(a,b,c).

3.3.2: Operations on Languages

Let L and M be two languages over the same alphabet A.

Definition: The union of L and M, written L ∪ M, is the set-theoretic union, i.e., it consists of all words (strings) in either L or M (or both).

Example: Let the alphabet be ascii. The union of {Grammatical English sentences with one, three, or five words} with {Grammatical English sentences with two or four words} is {Grammatical English sentences with five or fewer words}.

Definition: The concatenation of L and M is the set of all strings st, where s is a string of L and t is a string of M.

We again view concatenation as a product and write LM for the concatenation of L and M.

Examples:: Let the alphabet A={a,b,c,1,2,3}. The concatenation of the languages L={a,b,c} and M={1,2} is L∪M={a1,a2,b1,b2,c1,c2}. The concatenation of {aa,b,c} and {1,2,ε} is {aa1,aa2,aa,b1,b2,b,c1,c2,c}.

Definition: As with strings, it is natural to define powers of a language L.
L⁰={ε}, which is not φ.
Lⁱ⁺¹=LⁱL.

Definition: The (Kleene) closure of L, denoted L^* is
L⁰ ∪ L¹ ∪ L² ...

Definition: The positive closure of L, denoted L⁺ is
L¹ ∪ L² ...

Note: Given either closure it is easy to get the other one.

• L^* = {ε} ∪ L⁺
• L⁺ = L L^*

Example: Let both the alphabet and the language L be {0,1,2,3,4,5,6,7,8,9}. More formally, I would say the alphabet A is {0,1,2,3,4,5,6,7,8,9} and the language L is the set of all strings of length one over A. L⁺ gives all unsigned integers, but with some ugly versions. It has 3, 03, 000003.
{0} ∪ ( {1,2,3,4,5,6,7,8,9} ({0,1,2,3,4,5,6,7,8,9}^* ) ) seems better.

In these notes I may write * for ^* and + for ⁺, but that is strictly speaking wrong and I will not do it on the board or on exams or on lab assignments.

Example: {a,b}* is {ε,a,b,aa,ab,ba,bb,aaa,aab,aba,abb,baa,bab,bba,bbb,...}.
{a,b}+ is {a,b,aa,ab,ba,bb,aaa,aab,aba,abb,baa,bab,bba,bbb,...}.
{ε,a,b}* is {ε,a,b,aa,ab,ba,bb,...}.
{ε,a,b}+ is the same as {ε,a,b}*.

The book gives other examples based on L={letters} and D={digits}, which you should read.

3.3.3: Regular Expressions

The idea is that the regular expressions over an alphabet consist of
the alphabet, and expressions using union, concatenation, and *,
but it takes more words to say it right. For example, I didn't include (). Note that (A ∪ B)* is definitely not A* ∪ B* (* does not distribute over ∪) so we need the parentheses.

The book's definition includes many () and is more complicated than I think is necessary. However, it has the crucial advantages of being correct and precise.

The wikipedia entry doesn't seem to be as precise.

I will try a slightly different approach, but note again that there is nothing wrong with the book's approach (which appears in both first and second editions, essentially unchanged).

Parentheses, if present, control the order of operations. Without parentheses the following precedence rules apply.

The postfix unary operator * has the highest precedence. The book mentions that it is left associative. (I don't see how a postfix unary operator can be right associative or how a prefix unary operator such as unary minus could be left associative.)

Concatenation has the second highest precedence and is left associative.

| has the lowest precedence and is left associative.

The book gives various algebraic laws (e.g., associativity) concerning these operators.

As mentioned above, we don't need the positive closure since, for any RE,

    r+ = rr*.
  

Homework: 2(a-d), 4.

Examples
Let the alphabet A={a,b,c}. Write a regular expression representing the language consisting of all words with

1. at least one c.
2. every b followed by a c.
3. the same number of b's as c's

1. (a|b|c)*c(a|b|c)*
2. (a|bc|c)*
3. IMPOSSIBLE, regular expressions can't count (discussed later in the course)!

Start Lecture #4

3.3.4: Regular Definitions

      d1 → r1
      d2 → r2
      ...
      dn → rn
    

Example: C identifiers can be described by the following regular definition

    letter_ → A | B | ... | Z | a | b | ... | z | _
    digit → 0 | 1 | ... | 9
    CId → letter_ ( letter_ | digit)*
  

Regular definitions are just a convenience; they add no power to regular expressions. The C identifier example can be done simply as a regular expression by simply plugging in the earlier definitions to the later ones.

3.3.5: Extensions of Regular Expressions

There are many extensions of the basic regular expressions given above. The following three, especially the third, will be occasionally used in this course as they are useful for lexical analyzers.

All three are simply shorthand. That is, the set of possible languages generated using the extensions is the same as the set of possible languages generated without using the extensions.

1. One or more instances. This is the positive closure operator + mentioned above.
2. Zero or one instance. The unary postfix operator ? defined by
   r? = r | ε for any RE r.
3. Character classes. If a₁, a₂, ..., a_n are symbols in the alphabet, then
   [a₁a₂...a_n] = a₁ | a₂ | ... | a_n. In the special case where all the a's are consecutive, we can simplify the notation further to just [a₁-a_n].

Examples:

1. C-language identifiers

   	letter_ → [A-Za-z_]
   	digit → [0-9]
   	CId → letter_ ( letter_ | digit )*
         

2. Unsigned integer or floating point numbers

   	digit → [0-9]
   	digits → digit+
   	number → digits (. digits)?(E[+-]? digits)?
         

   This is actually fairly restrictive. For example, it doesn't permit 3. as a number.

Homework: 1(a) (you might need to read a C manual first to find out all the numerical constants in C).

3.4: Recognition of Tokens

Our current goal is to perform the lexical analysis needed for the following grammar.

    stmt → if expr then stmt
         | if expr then stmt else stmt
         | ε
    expr → term relop term   // relop is relational operator =, >, etc
         | term
    term →  id
         | number
  

Recall that the terminals are the tokens, the nonterminals produce terminals.

A regular definition for the terminals is

    digit → [0-9]
    digits → digits+
    number → digits (. digits)? (E[+-]? digits)?
    letter → [A-Za-z]
    id → letter ( letter | digit )*
    if → if
    then → then
    else → else
    relop → < | > | <= | >= | = | <>
  

On the board show how this can be done with just REs.

We also want the lexer to remove whitespace so we define a new token

    ws → ( blank | tab | newline ) +
  

Recall that the lexer will be called by the parser when the latter needs a new token. If the lexer then recognizes the token ws, it does not return it to the parser but instead goes on to recognize the next token, which is then returned. Note that you can't have two consecutive ws tokens in the input because, for a given token, the lexer will match the longest lexeme starting at the current position that yields this token. The table on the right summarizes the situation.

For the parser all the relational ops are to be treated the same so they are all the same token, relop. Naturally, other parts of the compiler, for example the code generator, will need to distinguish between the various relational ops so that appropriate code is generated. Hence, they have distinct attribute values.

3.4.1: Transition Diagrams

A transition diagram is similar to a flowchart for (a part of) the lexer. We draw one for each possible token. It shows the decisions that must be made based on the input seen. The two main components are circles representing states (think of them as decision points of the lexer) and arrows representing edges (think of them as the decisions made).

The transition diagram (3.13) for relop is shown on the right.

1. The double circles represent accepting or final states at which point a lexeme has been found. There is often an action to be done (e.g., returning the token), which is written to the right of the double circle.
2. If we have moved one (or more) characters too far in finding the token, one (or more) stars are drawn.
3. An imaginary start state exists and has an arrow coming from it to indicate where to begin the process.

It is fairly clear how to write code corresponding to this diagram. You look at the first character, if it is <, you look at the next character. If that character is =, you return (relop,LE) to the parser. If instead that character is >, you return (relop,NE). If it is another character, return (relop,LT) and adjust the input buffer so that you will read this character again since you have not used it for the current lexeme. If the first character was =, you return (relop,EQ).

3.4.2: Recognition of Reserved Words and Identifiers

The next transition diagram corresponds to the regular definition given previously.

Note again the star affixed to the final state.

Two questions remain.

1. How do we distinguish between identifiers and keywords such as then, which also match the pattern in the transition diagram?
2. What is (gettoken(), installID())?

We will continue to assume that the keywords are reserved, i.e., may not be used as identifiers. (What if this is not the case—as in Pl/I, which had no reserved words? Then the lexer does not distinguish between keywords and identifiers and the parser must.)

We will use the method mentioned last chapter and have the keywords installed into the identifier table prior to any invocation of the lexer. The table entry will indicate that the entry is a keyword.

installID() checks if the lexeme is already in the table. If it is not present, the lexeme is installed as an id token. In either case a pointer to the entry is returned.

gettoken() examines the lexeme and returns the token name, either id or a name corresponding to a reserved keyword.

The text also gives another method to distinguish between identifiers and keywords.

3.4.3: Completion of the Running Example

So far we have transition diagrams for identifiers (this diagram also handles keywords) and the relational operators. What remains are whitespace, and numbers, which are respectively the simplest and most complicated diagrams seen so far.

Recognizing Whitespace

The diagram itself is quite simple reflecting the simplicity of the corresponding regular expression.

• The delim in the diagram represents any of the whitespace characters, say space, tab, and newline.
• The final star is there because we needed to find a non-whitespace character in order to know when the whitespace ends and this character begins the next token.
• There is no action performed at the accepting state. Indeed the lexer does not return to the parser, but starts again from its beginning as it still must find the next token.

Recognizing Numbers

This certainly looks formidable, but it is not that bad; it follows from the regular expression.

In class go over the regular expression and show the corresponding parts in the diagram.

When an accepting states is reached, action is required but is not shown on the diagram. Just as identifiers are stored in a identifier table and a pointer is returned, there is a corresponding number table in which numbers are stored. These numbers are needed when code is generated. Depending on the source language, we may wish to indicate in the table whether this is a real or integer. A similar, but more complicated, transition diagram could be produced if the language permitted complex numbers as well.

Homework: 1 (only the ones done before).

3.4.4: Architecture of a Transition-Diagram-Based Lexical Analyzer

The idea is that we write a piece of code for each decision diagram. I will show the one for relational operations below. This piece of code contains a case for each state, which typically reads a character and then goes to the next case depending on the character read. The numbers in the circles are the names of the cases.

Accepting states often need to take some action and return to the parser. Many of these accepting states (the ones with stars) need to restore one character of input. This is called retract() in the code.

What should the code for a particular diagram do if at one state the character read is not one of those for which a next state has been defined? That is, what if the character read is not the label of any of the outgoing arcs? This means that we have failed to find the token corresponding to this diagram.

The code calls fail(). This is not an error case. It simply means that the current input does not match this particular token. So we need to go to the code section for another diagram after restoring the input pointer so that we start the next diagram at the point where this failing diagram started. If we have tried all the diagram, then we have a real failure and need to print an error message and perhaps try to repair the input.

Note that the order the diagrams are tried is important. If the input matches more than one token, the first one tried will be chosen.

    TOKEN getRelop()                        // TOKEN has two components
      TOKEN retToken = new(RELOP);          // First component set here
      while (true)
         switch(state)
           case 0: c = nextChar();
                   if (c == '<')      state = 1;
                   else if (c == '=') state = 5;
                   else if (c == '>') state = 6;
                   else fail();
                   break;
           case 1: ...
           ...
           case 8: retract();  // an accepting state with a star
                   retToken.attribute = GT;  // second component
                   return(retToken);
  

Alternate Methods

The book gives two other methods for combining the multiple transition-diagrams (in addition to the one above).

1. Unlike the method above, which tries the diagrams one at a time, the first new method tries them in parallel. That is, each character read is passed to each diagram (that hasn't already failed). Care is needed when one diagram has accepted the input, but others still haven't failed and may accept a longer prefix of the input.
2. The final possibility discussed, which appears to be promising, is to combine all the diagrams into one. That is easy for the example we have been considering because all the diagrams begin with different characters being matched. Hence we just have one large start with multiple outgoing edges. It is more difficult when there is a character that can begin more than one diagram.

3.5: The Lexical Analyzer Generator Lex

We are skipping 3.5 because

1. We will be writing our lexer from scratch.
2. What is here is not enough to learn how to use lex/flex.
3. If you are interested in learning how to use them you need to read (at least) the manual.

      declarations
      %%
      translation rules
      %%
      auxiliary functions
    

      %{
          /* definitions of manifest constants
             LT, LE, EQ, NE, GT, GE,
             IF, THEN, ELSE, ID, NUMBER, RELOP */
      %}

      /* regular definitions */
      delim     [ \t\n]
      ws        {delim}*
      letter    [A-Za-z]
      digit     [0-9]
      id        {letter}({letter}{digit})*
      number    {digit}+(\.{digit}+)?(E[+-]?{digit}+)?

      %%

      {ws}      {/* no action and no return */}
      if        {return(IF);}
      then      {return(THEN);}
      else      {return(ELSE);}
      {id}      {yylval = (int) installID(); return(ID);}
      {number}  {yylval = (int) installNum(); return(NUMBER);}
      "<"       {yylval = LT; return(RELOP);}
      "<="      {yylval = LE; return(RELOP);}
      "="       {yylval = EQ; return(RELOP);}
      "<>"      {yylval = NE; return(RELOP);}
      ">"       {yylval = GT; return(RELOP);}
      ">="      {yylval = GE; return(RELOP);}

      %%

      int installID() {/* function to install the lexeme, whose first
	                  character is pointed to by yytext, and whose
                          length is yyleng, into the symbol table and
                          return a pointer thereto */
      }

      int installNum() {/* similar to installID, but puts numerical
	                   constants into a separate table */
    

      #define LT 12
      #define LE 13
    

      IF / \(.*\){letter}
    

3.6: Finite Automata

The secret weapon used by lex et al to convert (compile) its input into a lexer.

Finite automata are like the graphs we saw in transition diagrams but they simply decide if a sentence (input string) is in the language (generated by our regular expression). That is, they are recognizers of the language.

There are two types of finite automata

1. Deterministic finite automata (DFA) have for each state (circle in the diagram) exactly one edge leading out for each symbol. So if you know the next symbol and the current state, the next state is determined. That is, the execution is deterministic; hence the name.
2. Nondeterministic finite automata (NFA) are the other kind. There are no restrictions on the edges leaving a state: there can be several with the same symbol as label and some edges can be labeled with ε. Thus there can be several possible next states from a given state and a current lookahead symbol.

Surprising Theorem: Both DFAs and NFAs are capable of recognizing the same languages, the regular languages, i.e., the languages generated by regular expressions (plus the automata can recognize the empty language).

What Does This Mean?

There are certainly NFAs that are not DFAs. But the language recognized by each such NFA can also be recognized by at least one DFA.

Why Mention (Confusing) NFAs?

The DFA that recognizes the same language as an NFA might be significantly larger that the NFA.

The finite automaton that one constructs naturally from a regular expression is often an NFA.

3.6.1: Nondeterministic Finite Automata

Here is the formal definition.

A nondeterministic finite automaton (NFA) consists of

1. A finite set of states S.
2. An input alphabet Σ not containing ε.
3. A transition function that gives, for each state and each symbol in Σ ∪ ε, a set of next states (or successor states).
4. An element s₀ of S, the start state.
5. A subset F of S, the accepting states (or final states).

An NFA is basically a flow chart like the transition diagrams we have already seen. Indeed an NFA (or a DFA, to be formally defined soon) can be represented by a transition graph whose nodes are states and whose edges are labeled with elements of Σ ∪ ε. The differences between a transition graph and our previous transition diagrams are:

1. Possibly multiple edges with the same label leaving a single state.
2. An edge may be labeled with ε.

The transition graph to the right is an NFA for the regular expression (a|b)^*abb, which (given the alphabet {a,b}) represents all words ending in abb.

Consider aababb. If you choose the wrong edge for the initial a's you will get stuck. But an NFA accepts a word if any path (beginning at the start state and using the symbols in the word in order) ends at an accepting state.

It essentially tries all such paths at once and accepts if any end at an accepting state. This means in addition that, if some paths uses all the input and end in a non-accepting state, but at least one path uses all the input (doesn't get stuck) and ends in an accepting state, the input is accepted.

Remark: Patterns like (a|b)*abb are useful regular expressions! If the alphabet is ascii, consider *.java.

Homework: 3, 4.

3.6.2: Transition Tables

There is an equivalent way to represent an NFA, namely a table giving, for each state s and input symbol x (and ε), the set of successor states x leads to from s. The empty set φ is used when there is no edge labeled x emanating from s. The table on the right corresponds to the transition graph above.

The downside of these tables is their size, especially if most of the entries are φ since those entries would not take any space in a transition graph.

Homework: 5.

3.6.3: Acceptance of Input Strings by Automata

An NFA accepts a string if the symbols of the string specify a path from the start to an accepting state.

Again note that these symbols may specify several paths, some of which lead to accepting states and some that don't. In such a case the NFA does accept the string; one successful path is enough.

Also note that if an edge is labeled ε, then it can be taken for free.

For the transition graph above any string can just sit at state 0 since every possible symbol (namely a or b) can go from state 0 back to state 0. So every string can lead to a non-accepting state, but that is not important since if just one path with that string leads to an accepting state, the NFA accepts the string.

The language defined by an NFA or the language accepted by an NFA is the set of strings (a.k.a. words) accepted by the NFA.

So the NFA in the diagram above accepts the same language as the regular expression (a|b)*abb.

If A is an automaton (NFA or DFA) we use L(A) for the language accepted by A.

The diagram on the right illustrates an NFA accepting the language L(aa*|bb*). The path
0 → 3 → 4 → 4 → 4 → 4
shows that bbbb is accepted by the NFA.

Note how the ε that labels the edge 0 → 3 does not appear in the string bbbb since ε is the empty string.

3.6.4: Deterministic Finite Automata

There is something weird about an NFA if viewed as a model of computation. How is a computer of any realistic construction able to check out all the (possibly infinite number of) paths to determine if any terminate at an accepting state?

We now consider a much more realistic model, a DFA.

Definition: A deterministic finite automata or DFA is a special case of an NFA having the restrictions

1. No edge is labeled with ε
2. For any state s and symbol a, there is exactly one edge leaving s with label a. (If no edge is shown with label a, there is an implied edge leading to a node labeled fail.)

This is realistic. We are at a state and examine the next character in the string, depending on the character we go to exactly one new state. Looks like a switch statement to me.

Minor point: when we write a transition table for a DFA, the entries are elements not sets so there are no {} present.

Simulating a DFA

Indeed a DFA is so reasonable there is an obvious algorithm for simulating it (i.e., reading a string and deciding whether or not it is in the language accepted by the DFA). We present it now.

    s := s0;   // start state.
    c := nextChar();      // a priming read
    while (c /= eof) {
      s := move(s,c);
      c := nextChar();
    }
    if (s is in F, the set of accepting states) return yes
    else return no
  

3.7: From Regular Expressions to Automata

3.7.0: Not Losing Site of the Forest Due to the Trees

This is not from the book.

Do not forget the goal of the chapter is to understand lexical analysis. (Languages generated by) regular expressions are the key to this task, since we use them to specify tokens. So we want to recognize (languages generated by) regular expressions. We are going to see two methods.

1. Convert the regular expression to an NFA and simulate the NFA.
2. Convert the regular expression to an NFA, convert the NFA to a DFA, and simulate the DFA.

So we need to learn 4 techniques.

1. Convert a regular expression to an NFA
2. Simulate an NFA
3. Convert an NFA to a DFA
4. Simulate a DFA.

The list I just gave is in the order the algorithms would be applied—but you would use either 2 or (3 and 4).

However, we will follow the order in the book, which is exactly the reverse.

Indeed, we just did item #4 and will now do #3.

3.7.1: Converting an NFA to a DFA

The book gives a detailed proof; I am just trying to motivate the ideas.

Let N be an NFA, we construct D, a DFA that accepts the same strings as does N. Call a state of N an N-state, and call a state of D a D-state.

The idea is that a D-state corresponds to a set of N-states and hence the procedure we are describing is called the subset algorithm. Specifically for each string X of symbols we consider all the N-states that can result when N processes X. This set of N-states is a D-state. Let us consider the transition graph on the right, which is an NFA that accepts strings satisfying the regular expression
(a|b)^*abb. The alphabet is {a,b}.

The start state of D is the set of N-states that can result when N processes the empty string ε. This is called the ε-closure of the start state s₀ of N, and consists of those N-states that can be reached from s₀ by following edges labeled with ε. Specifically it is the set {0,1,2,4,7} of N-states. We call this state D₀ and enter it in the transition table we are building for D on the right.

Next we want the a-successor of D₀, i.e., the D-state that occurs when we start at D₀ and move along an edge labeled a. We call this successor D₁. Since D₀ consists of the N-states corresponding to ε, D₁ is the N-states corresponding to εa=a. We compute the a-successor of all the N-states in D₀ and then form the ε-closure.

Next we compute the b-successor of D₀ the same way and call it D₂.

We continue forming a- and b-successors of all the D-states until no new D-states result (there are only a finite number of subsets of all the N-states so this process does indeed stop).

This gives the table on the right. D₄ is the only D-accepting state as it is the only D-state containing the (only) N-accepting state 10.

Theoretically, this algorithm is awful since for a set with k elements, there are 2^k subsets. Fortunately, normally only a small fraction of the possible subsets occur in practice. For example, the NFA we just did had 11 states so there are 2048 possible subsets, only 5 of which occurred in the DFA we constructed.

Homework: 1.

3.7.2: Simulating an NFA

Instead of producing the DFA, we can run the subset algorithm as a simulation itself. This is item #2 in my list of techniques

    S = ε-closure(s0);
    c = nextChar();
    while ( c != eof ) {
      S = ε-closure(move(S,c));
      c = nextChar();
    }
    if ( S ∩ F != φ ) return yes;   // F is accepting states
    else return no;
  

Homework: 2.

3.7.3: Efficiency of NFA Simulation

Slick implementation.

3.7.4: Constructing an NFA from a Regular Expression

I give a pictorial proof by induction. This is item #1 from my list of techniques.

1. The base cases are the empty regular expression and the regular expression consisting of a single symbol a in the alphabet.
2. The inductive cases are.
   1. s | t for s and t regular expressions
   2. st for s and t regular expressions
   3. s^*
   4. (s), which is trivial since the nfa for s also works for (s).
3. We have so few cases because regular expressions (w/o extensions) are sufficient. That is, once we can construct arbitrary REs, we can also construct the extended REs and the regular definitions since both of these can be expressed as REs.

The pictures on the right illustrate the base and inductive cases.

Remarks:

1. The generated NFA has at most twice as many states as there are operators and operands in the RE. This is important for studying the complexity of the NFA, which we will not do.
2. The generated NFA has one start and one accepting state. The accepting state has no outgoing arcs and the start state has no incoming arcs. This is important for the pictorial proof we gave since we always assumed that the constituent NFAs had that property when building the bigger NFS.
3. Note that the diagram for st correctly indicates that the final state of s and the initial state of t are merged. This is one use of the previous remark that there is only one start state and one final state.
4. Except for the accepting state, each state of the generated NFA has either one outgoing arc labeled with a symbol or two outgoing arcs labeled with ε.

Do the NFA for (a|b)^*abb and see that we get the same diagram that we had before.

Do the steps in the normal leftmost, innermost order (or draw a normal parse tree and follow it).

Homework: 3 a,b,c

3.8: Design of a Lexical-Analyzer Generator

How lexer-generators like Lex work. Since your lab2 is to produce a lexer, this is also a section on how you should solve lab2.

3.8.1: The structure of the generated analyzer

We have seen simulators for DFAs and NFAs.

Recall that the regular definitions in Lex are mere conveniences that can easily be converted to REs and hence we need only convert REs into an FSA.

We already know how to convert a single RE into an NFA. But lex input will contain several REs (since it wishes to recognize several different tokens). The solution is to

1. Produce an NFA for each RE.
2. Introduce a new start state.
3. Introduce an ε transition from the new start state to the start of each NFA constructed in step 1.
4. When one of the NFAs reaches one of the accepting states, the simulation does NOT stop. See below for an explanation.

Label each of the accepting states (for all NFAs constructed in step 1) with the actions specified in the lex program for the corresponding pattern.

3.8.2: Pattern Matching Based on NFAs

We use the algorithm for simulating NFAs presented in 3.7.2.

The simulator starts reading characters and calculates the set of states it is at.

At some point the input character does not lead to any state or we have reached the eof. Since we wish to find the longest lexeme matching the pattern we proceed backwards from the current point (where there was no state) until we reach an accepting state (i.e., the set of NFA states, N-states, contains an accepting N-state). Each accepting N-state corresponds to a matched pattern. The lex rule is that if a lexeme matches multiple patterns we choose the pattern listed first in the lex-program.

Example

Consider the example on the right with three patterns and their associated actions and consider processing the input aaba.

1. We begin by constructing the three NFAs. To save space, the third NFA is not the one that would be constructed by our algorithm, but is an equivalent smaller one. For example, some unnecessary ε-transitions have been eliminated. If one views the lex executable as a compiler transforming lex source into NFAs, this would be considered an optimization.
2. We introduce a new start state and ε-transitions as in the previous section.
3. We start at the ε-closure of the start state, which is {0,1,3,7}.
4. The first a (remember the input is aaba) takes us to {2,4,7}. This includes an accepting state and indeed we have matched the first patten. However, we do not stop since we may find a longer match.
5. The next a takes us to {7}.
6. The b takes us to {8}.
7. The next a fails since there are no a-transitions out of state 8. So we must back up to before trying the last a.
8. We are back in {8} and ask if one of these N-states (I know there is only one, but there could be more) is an accepting state.
9. Indeed state 8 is accepting for third pattern. If there were more than one accepting state in the list, we would choose the one in the earliest listed pattern.
10. Action3 would now be performed.

3.8.3: DFA's for Lexical Analyzers

We could also convert the NFA to a DFA and simulate that. The resulting DFA is on the right. Note that it shows the same D-states (i.e., sets of N-states) we saw in the previous section, plus some other D-states that arise from inputs other than aaba.

We label the accepting states with the pattern matched. If multiple patterns are matched (because the accepting D-state contains multiple accepting N-states), we use the first pattern listed (assuming we are using lex conventions). For example, the middle D-state on the bottom row contains two accepting N-states, 6 and 8. Since the RE for 6 was listed first, it appears below the state.

Consider processing the string aa. Show how you get two tokens.

Technical point. For a DFA, there must be a outgoing edge from each D-state for each possible character. In the diagram, when there is no NFA state possible, we do not show the edge. Technically we should show these edges, all of which lead to the same D-state, called the dead state, and corresponds to the empty subset of N-states.

Remark: A pure DFA or NFA simply accepts or rejects a single string. In the context of lexers, that would mean accepting or rejecting a single lexeme. But a lexer isn't given a single lexeme to process, it is given a program consisting of many lexemes. That is why we have the complication of backtracking after getting stuck; a pure NFA/DFA would just say reject if the next character did not correspond to an outgoing arc.

Alternatives for Implementing Lab 2

There are trade-offs depending on how much you want to do by hand and how much you want to program. At the extreme you could write a program that reads in the regular expression for the tokens and produces a lexer, i.e., you could write a lexical-analyzer-generator. I very much advise against this, especially since the first part of the lab requires you to draw the transition diagrams anyway.

The two reasonable alternatives are.

1. By hand, convert the NFA to a DFA and then write your lexer based on this DFA, simulating its actions for input strings.
2. Write your program based on the NFA.

Remark: Lab 2 assigned. It is due in 2 weeks, 9 October 2008.>

Start Lecture #5

Chapter 4: Syntax Analysis

Homework: Read Chapter 4.

4.1: Introduction

4.1.1: The role of the parser

As we saw in the previous chapter the parser calls the lexer to obtain the next token.

Conceptually, the parser accepts a sequence of tokens and produces a parse tree. In practice this might not occur.

1. The source program might have errors. Shamefully, we will do very little error handling.
2. Instead of explicitly constructing the parse tree, the actions that the downstream components of the front end would do on the tree can be integrated with the parser and done incrementally on components of the tree. We will see examples of this, but your lab number 3 will produce a parse tree. Your lab number 4 will process this parse tree and do the actions.
3. Real compilers produce (abstract) syntax trees not parse trees (concrete syntax trees). We don't do this for the pedagogical reasons given previously.

There are three classes for grammar-based parsers.

1. universal
2. top-down
3. bottom-up

The universal parsers are not used in practice as they are inefficient; we will not discuss them.

As expected, top-down parsers start from the root of the tree and proceed downward; whereas, bottom-up parsers start from the leaves and proceed upward.

The commonly used top-down and bottom parsers are not universal. That is, there are context-free grammars that cannot be used with them.

The LL (top down) and LR (bottom-up) parsers are important in practice. Hand written parsers are often LL. Specifically, the predictive parsers we looked at in chapter two are for LL grammars.

The LR grammars form a larger class. Parsers for this class are usually constructed with the aid of automatic tools.

4.1.2: Representative Grammars

Expressions with + and *

    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  

This takes care of precedence, but as we saw before, gives predictive parsing trouble since it is left-recursive. So we used the following non-left-recursive grammar that generates the same language.

    E  → T E'
    E' → + T E' | ε
    T  → F T'
    T' → * F T' | ε
    F  → ( E ) | id
  

The following ambiguous grammar will be used for illustration, but in general we try to avoid ambiguity.

    E → E + E | E * E | ( E ) | id
  

4.1.3: Syntax Error Handling

There are different levels of errors.

1. Lexical errors: For example, spelling.
2. Syntactic errors: For example missing ; .
3. Semantic errors: For example wrong number of array indexes.
4. Logical errors: For example off by one usage of < instead of <=.

4.1.4: Error-Recovery Strategies

The goals are clear, but difficult.

• Report errors clearly and accurately. One difficulty is that one error can mask another and can cause correct code to look faulty.
• Recover quickly enough to not miss other errors.
• Add minimal overhead.

Trivial Approach: No Recovery

Print an error message when parsing cannot continue and then terminate parsing.

Panic-Mode Recovery

The first level improvement. The parser discards input until it encounters a synchronizing token. These tokens are chosen so that the parser can make a fresh beginning. Good examples for C/Java are ; and }.

Phrase-Level Recovery

Locally replace some prefix of the remaining input by some string. Simple cases are exchanging ; with , and = with ==. Difficulties occur when the real error occurred long before an error was detected.

Error Productions

Include productions for common errors.

Global Correction

Change the input I to the closest correct input I' and produce the parse tree for I'.

4.2: Context-Free Grammars

4.2.1: Formal Definition

4.2.2: Notational Conventions

I am not as formal as the book. In particular, I don't use italics. Nonetheless I do (try to) use some of the conventions, in particular the ones below. Please correct me if I violate them.

• Uppercase letters early in the alphabet like A, B, and C are use to represent nonterminals. Also S is often used as the start symbol, but remember that if no explicit start symbol is given for a grammar, than the LHS of the first production is the start symbol.
• Lowercase letters early in the alphabet are used to represent terminals.
• Uppercase letters late in the alphabet are used to represent a single grammar symbol, i.e., either a single terminal or a single nonterminal.
• Lowercase letters late in the alphabet represent are used to represent a (possibly empty) string of terminals.
• Lowercase Greek letters are used to represent a (possibly empty) string of grammar symbols. I never use ε for this purpose since ε always represents the empty string.
• So A → α would be a generic production in a context-free grammar.
• X → α and β → α include illegal productions for a context-free grammar.

As I have mentioned before, when the entire grammar is written, no conventions are needed to tell the nonterminals, terminals, and start symbol. The nonterminals are the LHSs, the terminals are everything else on the RHS, and the start symbol is the LHS of the first production. The the notational conventions are used when you just give a few productions not a full grammar.

4.2.3: Derivations

This is basically just notational convenience, but important nonetheless.

Assume we have a production A → α. We would then say that A derives α and write
A ⇒ α

We generalize this. If, in addition, β and γ are strings (each may contain terminals and/or nonterminals), we say that βAγ derives βαγ and write

    βAγ ⇒ βαγ
  

We generalize further. If α derives β and β derives γ, we say α derives γ and write
α ⇒* γ.

The notation used is ⇒ with a * over it (I don't see it in html). This should be read derives in zero or more steps. Formally,

1. α ⇒* α, for any string α.
2. If α ⇒* β and β ⇒ γ, then α ⇒* γ.

A sentential form may contain nonterminals and terminals.

Examples: Recall the ambiguous grammar above

    E → E + E | E * E | ( E ) | id
  

    E ⇒ E + E ⇒ id + E ⇒ id + id
    E ⇒ E + E ⇒ E + id ⇒ id + id
  

(Since both derivations give the same parse tree, this does not show the grammar is ambiguous. You should be able to find a sentence—without looking back in the notes—that has two different parse trees.)

In the first derivation shown just above, each step replaced the leftmost nonterminal by the body of a production having the nonterminal as head. This is called a leftmost derivation. Similarly the second derivation, in which each step replaced the rightmost nonterminal, is called a rightmost derivation. Sometimes the latter are called canonical derivations, but we won't do so.

When one wishes to emphasize that a (one step) derivation is leftmost they write an lm under the ⇒. To emphasize that a (general) derivation is leftmost, one writes an lm under the ⇒*. Similarly one writes rm to indicate that a derivation is rightmost. I won't do this in the notes but will on the board.

Homework: 1(ab), 2(ab).

4.2.4: Parse Trees and Derivations

The leaves of a parse tree (or of any other tree), when read left to right, are called the frontier of the tree. For a parse tree we also call them the yield of the tree.

If you are given a derivation starting with a single nonterminal,
A ⇒ α₁ ⇒ α₂ ... ⇒ α_n it is easy to write a parse tree with A as the root and α_n as the leaves. Just do what (the production contained in) each step of the derivation says. The LHS of each production is a nonterminal in the frontier of the current tree so replace it with the RHS to get the next tree.

Do this for both the leftmost and rightmost derivations of id+id above.

So there can be many derivations that wind up with the same final tree.

But for any parse tree there is a unique leftmost derivation producing that tree (always choose the leftmost unmarked nonterminal to be the LHS, mark it, and write the production with this LHS and the children as the RHS). Similarly, there is a unique rightmost derivation that produces the tree. There may be others as well (e.g., sometime choose the leftmost unmarked nonterminal to expand and other times choose the rightmost; or choose a middle unmarked nonterminal).

Homework: 1c

4.2.5: Ambiguity

Recall that an ambiguous grammar is one for which there is more than one parse tree for a single sentence. Since each parse tree corresponds to exactly one leftmost derivation, a grammar is ambiguous if and only if it permits more than one leftmost derivation of a given sentence. Similarly, a grammar is ambiguous if and only if it permits more than one rightmost of a given sentence.

We know that the grammar

    E → E + E | E * E | ( E ) | id
  

    id + id * id
  

    E ⇒ E + E          E ⇒ E * E
      ⇒ id + E           ⇒ E + E * E
      ⇒ id + E * E       ⇒ id + E * E
      ⇒ id + id * E      ⇒ id + id * E
      ⇒ id + id * id     ⇒ id + id * id
  

As we stated before we prefer unambiguous grammars. Failing that, we want disambiguation rules, as are often given for the dangling else in the C language.

4.2.7: Context-Free Grammars Versus Regular Expressions

Alternatively context-free languages vs regular languages.

Given an RE, construct an NFA as in chapter 3.

From that NFA construct a grammar as follows.

1. Define a nonterminal A_i for each state i.
2. For a transition from A_i to A_j on input a (or ε), add a production
   A_i → aA_j (or A_i → A_j).
3. If i is accepting, add A_i → ε
4. If i is start, make A_i start.

If you trace an NFA accepting a sentence, it just corresponds to the constructed grammar deriving the same sentence. Similarly, follow a derivation and notice that at any point prior to acceptance there is only one nonterminal; this nonterminal gives the state in the NFA corresponding to this point in the derivation.

The book starts with (a|b)*abb and then uses the short NFA on the left below. Recall that the NFA generated by our construction is the longer one on the right.

The book gives the simple grammar for the short diagram.

Let's be ambitious and try the long diagram

    A0 → A1 | A7
    A1 → A2 | A4
    A2 → a A3
    A3 → A6
    A4 → b A5
    A5 → A6
    A6 → A1 | A7
    A7 → a A8
    A8 → b A9
    A9 → b A10
    A10 → ε
  

Now trace a path in the NFA beginning at the start state and see that it is just a derivation. That is the string corresponding to that path is a sentential form. The same is true in reverse (derivation gives path). The key is that at every stage you have at only one nonterminal.

Then notice that when you get to an accepting state, you have no nonterminals so accepting a string in the NFA shows it is a sentence in the language

Grammars, but not Regular Expressions, Can Count

The grammar

    A → c A b | ε
  

A proof is in the book. The idea is that you need a infinite number of states to represent the number of c's you have seen so that you can ensure that you see the same number of b's. But a RE is equivalent to a DFA (or NFA) and the F stands for finite.

4.3: Writing a Grammar

4.3.1: Lexical vs Syntactic Analysis

Why have a separate lexer and parser? Since the lexer deals with REs / Regular Languages and the parser deals with the more powerful Context Free Grammars (CFGs) / Context Free Languages (CFLs), everything a lexer can do, a parser could do as well. The reasons for separating the lexer and parser are from software engineering considerations.

• REs are easier than CFGs to understand.
• More efficient algorithms/tools exist for automating the RE-based lexer than for the CFG-based parser.
• Only the lexer need deal with the external environment.

      stmt → if expr then stmt
      | if expr then stmt else stmt
      | other
    

        stmt → matched-stmt | open-stmt
matched-stmt → if expr then matched-stmt else matched-stmt
	     | other
   open-stmt → if expr then stmt
	     | if expr then matched-stmt else open-stmt
    

4.3.3: Eliminating Left Recursion

We did special cases in chapter 2. Now we do it right(tm).

Previously we did it separately for one production and for two productions with the same nonterminal on the LHS. Not surprisingly, this can be done for n such productions (together with other non-left recursive productions involving the same nonterminal).

Specifically we start with

    A → A α1 | A α2 | ... A αn | β1 | β2 | ... βm
  

The equivalent non-left recursive grammar is

    A  → β1 A' | ... | βm A'
    A' → α1 A' | ... | αn A' | ε
  

The idea is as follows. Look at the left recursive grammar. At some point you stop producing more As and have the A (which is always on the left) become one of the βs. So the final string starts with a β. Up to this point all the As became Aα for one of the αs. So the final string is a β followed by a bunch of αs, which is exactly what the non-left recursive definition says.

Example: Assume n=m=1, α₁ is + and β₁ is *. So the left recursive grammar is

    A → A + | *
  

    A  → * A'
    A' → + A' | ε
  

    A ⇒ A + ⇒ A + + ⇒ * + +
  

    A ⇒ * A' ⇒ * + A' ⇒ * + + A' ⇒ * + +
  

This procedure removes direct left recursion where a production with A on the left hand side begins with A on the right. If you also had direct left recursion with B, you would apply the procedure twice.

The harder general case is where you permit indirect left recursion, where, for example one production has A as the LHS and begins with B on the RHS, and a second production has B on the LHS and begins with A on the RHS. Thus in two steps we can turn A into something starting again with A. Naturally, this indirection can involve more than 2 nonterminals.

Theorem: All left recursion can be eliminated.

Proof: The book proves this for grammars that have no ε-productions and no cycles and has exercises asking the reader to prove that cycles and ε-productions can be eliminated.

We will try to avoid these hard cases.

Homework: Eliminate left recursion in the following grammar for simple postfix expressions.

    S → S S + | S S * | a
  

4.3.4: Left Factoring

If two productions with the same LHS have their RHS beginning with the same symbol (terminal or nonterminal), then the FIRST sets will not be disjoint so predictive parsing (chapter 2) will be impossible and more generally top down parsing (defined later this chapter) will be more difficult as a longer lookahead will be needed to decide which production to use.

So convert A → α β1 | α β2 into

   A  → α A'
   A' → β1 | β2
  

Homework: Left factor your answer to the previous homework.

4.3.5: Non-CFL Constructs

Although our grammars are powerful, they are not all-powerful. For example, we cannot write a grammar that checks that all variables are declared before used.

4.4: Top-Down Parsing

We did an example of top down parsing, namely predictive parsing, in chapter 2.

For top down parsing, we

1. Start with the root of the parse tree, which is always the start symbol of the grammar. That is, initially the parse tree is just the start symbol.
2. Choose a nonterminal in the frontier.
   1. Choose a production having that nonterminal as LHS.
   2. Expand the tree by making the RHS the children of the LHS.
3. Repeat above until the frontier is all terminals.
4. Hope that the frontier equals the input string.

The above has two nondeterministic choices (the nonterminal, and the production) and requires luck at the end. Indeed, the procedure will generate the entire language. So we have to be really lucky to get the input string.

Another problem is that the procedure may not terminate.

4.4.1: Recursive Decent Parsing

Let's reduce the nondeterminism in the above algorithm by specifying which nonterminal to expand. Specifically, we do a depth-first (left to right) expansion. This corresponds to a leftmost derivation. That is, we expand the leftmost nonterminal in the frontier.

We leave the choice of production nondeterministic.

We also process the terminals in the RHS, checking that they match the input. By doing the expansion depth-first, left to right, we ensure that we encounter the terminals in the order they will appear in the frontier of the final tree. Thus if the terminal does not match the corresponding input symbol now, it never will (since there are no nonterminals to its left) and the expansion so far will not produce the input string as desired.

Now our algorithm is

1. Initially, the tree is the start symbol, the nonterminal we are currently processing.
   
   
2. Choose a production having the current nonterminal as LHS and expand the tree with the RHS, X1 X2 ... Xn.
   
   

3. for i = 1 to n
     if Xi is a nonterminal
       process Xi  // recursive call to step 2
     else if Xi (a terminal) matches current input symbol
       advance input to next symbol
     else // trouble Xi doesn't match and never will
         

Note that the trouble mentioned at the end of the algorithm does not signify an erroneous input. We may simply have chosen the wrong production in step 2.

In a general recursive descent (top-down) parser, we would support backtracking. That is, when we hit the trouble, we would go back and choose another production.

It is possible that no productions work for this nonterminal, because the wrong choice was made earlier. In that case we must back up further.

As an example consider the grammar

    S → A a | B b
    A → 1 | 2
    B → 1 | 2
  

The good news is that we will work with grammars where we can control the nondeterminism much better. Recall that for predictive parsing, the use of 1 symbol of lookahead made the algorithm fully deterministic, without backtracking.

4.4.2: FIRST and FOLLOW

We used FIRST(RHS) when we did predictive parsing.

Now we learn the whole truth about these two sets, which proves to be quite useful for several parsing techniques (and for error recovery, but we won't make use of this).

The basic idea is that FIRST(α) tells you what the first terminal can be when you fully expand the string α and FOLLOW(A) tells what terminals can immediately follow the nonterminal A.

Note that there might have been symbols between A and c during the derivation, providing they all derived ε and eventually c immediately follows A.

Unfortunately, the algorithms for computing FIRST and FOLLOW are not as simple to state as the definition suggests, in large part caused by ε-productions.

Calculating FIRST

Remember that FIRST(α) is the set of terminals that can begin a string derived from α. Since a terminal can derive only itself, FIRST of a terminal is trivial.

    FIRST(a) = {a}, for any terminal a
  

FIRST of a nonterminal is not so easy

1. Initialize FIRST(A)=φ for all nonterminals A
2. If A → ε is a production, add ε to FIRST(A).
3. For each production A → Y₁ ... Y_n,
   1. add to FIRST(A) any terminal a satisfying
      1. a is in FIRST(Y_i) and
      2. ε is in all previous FIRST(Y_j).
   2. add ε to FIRST(A) if ε is in all FIRST(Y_j).
   Repeat step 3 until nothing is added.

FIRST of an arbitrary string now follows.

1. FIRST of any string X=X₁X₂...X_n is initialized to φ.
2. add to FIRST(X) any non-ε symbol in FIRST(X_i) if ε is in all previous FIRST(X_j).
3. add ε to FIRST(X) if ε is in every FIRST(X_j). In particular if X is ε, FIRST(X)={ε}.

Calculating FOLLOW

Recall that FOLLOW is defined only for nonterminals A and is the set of terminals that can immediately follow A in a sentential form (a string derivable from the start symbol

1. Initialize FOLLOW(S)=$ and FOLLOW(A)=φ for all other nonterminals A.
2. For every production A → α B β, add all of FIRST(β) except ε to FOLLOW(B).
3. Apply the following rule until nothing is added to any FOLLOW set.
   For every production ending in B, i.e. for
       A → α B and for
       A → α B β, where FIRST(β) contains ε,
   add all of FOLLOW(A) to FOLLOW(B).

Do the FIRST and FOLLOW sets for

    E  → T E'
    E' → + T E' | ε
    T  → F T'
    T' → * F T' | ε
    F  → ( E ) | id
  

Start Lecture #6

Homework: Compute FIRST and FOLLOW for the postfix grammar S → S S + | S S * | a

4.4.3: LL(1) Grammars

The predictive parsers of chapter 2 are recursive descent parsers needing no backtracking. A predictive parser can be constructed for any grammar in the class LL(1). The two Ls stand for processing the input Left to right and for producing Leftmost derivations. The 1 in parens indicates that 1 symbol of lookahead is used.

The 2nd condition may seem strange; it did to me for a while. Let's consider the simplest case that condition 2 is trying to avoid. S is the start symbol

    A → ε      // β=ε so β derives ε
    A → c      // α=c so α derives a string beginning with c
    S → A c    // c is in FOLLOW(A)
  

Probably the simplest derivation possible is

    S ⇒ A c ⇒ c
  

Assume we are using predictive parsing and, as illustrated in the diagram to the right, we are at A in the parse tree and c in the input. Since lookahead=c and c is in FIRST(RHS) for the second A production, we would choose that production to expand A. But this is wrong! Remember that we don't look ahead in the tree, we look ahead just in the input. So we would not have noticed that the next node in the tree (i.e., in the frontier) is c. The next node can indeed be c since c is in FOLLOW(A). So we should have used the top A production to produce ε in the tree, and then the next node c would match the input c.

Constructing a Predictive Parsing Table

The goal is to produce a table telling us at each situation which production to apply. A situation means a nonterminal in the parse tree and an input symbol in lookahead.

So we produce a table with rows corresponding to nonterminals and columns corresponding to input symbols (including $, the endmarker). In an entry we put the production to apply when we are in that situation.

We start with an empty table M and populate it as follows. (2e has typo; it has FIRST(A) instead of FIRST(α).) For each production A → α

1. For each terminal a in FIRST(α), add A → α to M[A,a]. This is what we did with predictive parsing in chapter 2. The point was that if we are up to A in the tree and a is the lookahead, we could (should??) use the production A→α.
   
   
2. If ε is in FIRST(α), then add A → α to M[A,b] (resp. M[A,$]) for each terminal b in FOLLOW(A) (if $ is in FOLLOW(A)). This is not so obvious; it corresponds to the second (strange) condition above. If ε is in FIRST(α), then α⇒*ε. Hence we could (should??) apply the production A→α, have the α go to ε and then the b (or $) that follows A will match the b in the input.

When we have finished filling in the table M, what do we do if an slot has

1. no entries?
   This is a normal occurrence and does not indicate a problem with the table. Instead, it means that, if parsing gets to this situation, no production is appropriate. In other words parsing should never reference this table slot. Hence, if parsing an input does reference this entry, that input is not in the language generated by the grammar. A production quality parser might try to correct the input, but for us, it is enough to print an error message and give up.
   
   
2. one entry?
   Perfect! This means we know exactly what to do in this situation.
   
   
3. more than one entry?
   This should not happen since the present section is entitled LL(1) grammars. Mostly likely the problem is that the grammar is not LL(1). (Also possible is that an error was made in constructing the table.) Since the grammar is not LL(1), we must use a different technique instead of predictive parsing. One possibility is bottom-up parsing, which we study next. Another possibility is to modify the procedure for this nonterminal to look further ahead (typically one more token) to decide what action to perform.

Example: Work out the parsing table for

    E  → T E'
    E' → + T E' | ε
    T  → F T'
    T' → * F T' | ε
    F  → ( E ) | id
  

We already computed FIRST and FOLLOW as shown on the right. The table skeleton is

Example: What about ε-productions? Produce FIRST, FOLLOW, and the parsing table for

    S → B D
    B → b | ε
    D → d | ε
  

Homework: Produce the predictive parsing table for

1. S → 0 S 1 | 0 1
2. the prefix grammar S → + S S | * S S | a

Remark: Lab 3 will use the material up to here.

4.5: Bottom-Up Parsing

Now we start with the input string, i.e., the bottom (leaves) of what will become the parse tree, and work our way up to the start symbol.

For bottom up parsing, we are not fearful of left recursion as we were with top down. Our first few examples will use the left recursive expression grammar

    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  

4.5.1: Reductions

Remember that running a production in reverse, i.e., replacing the RHS by the LHS, is called reducing. So our goal is to reduce the input string to the start symbol.

On the right is a movie of parsing id*id in a bottom-up fashion. Note the way it is written. For example, from step 1 to 2, we don't just put F above id*id. We draw it as we do because it is the current top of the tree (really forest) and not the bottom that we are working on so we want the top to be in a horizontal line and hence easy to read.

The tops of the forest are the roots of the subtrees present in the diagram. For the movie those are

    id * id,  F * id,  T * id,  T * F,  T,  E
  

The steps from one frame of the movie, when viewed going down the page, are reductions (replace the RHS of a production by the LHS). Naturally, when viewed going up the page, we have a derivation (replace LHS by RHS). For our example the derivation is
E ⇒ T ⇒ T * F ⇒ T * id ⇒ F * id ⇒ id * id

Note that this is a rightmost derivation and hence each of the sets of roots identified above is a right sentential form. So the reduction we did in the movie was a rightmost derivation in reverse.

Remember that for a non-ambiguous grammar there is only one rightmost derivation and hence there is only one rightmost derivation in reverse.

Remark: You cannot simply scan the string (the roots of the forest) from left to right and choose the first substring that matches the RHS of some production. If you try it in our movie you will reduce T to E right after T appears. The result is not a right sentential form.

4.5.2: Handle Pruning

The strings that are reduced during the reverse of a rightmost derivation are called the handles. For our example, this is shown in the table on the right.

Note that the string to the right of the handle must contain only terminals. If there was a nonterminal to the right, it would have been reduced in the RIGHTmost derivation that leads to this right sentential form.

Often instead of referring to a derivation A→α as a handle, we call α the handle. I should say a handle because there can be more than one if the grammar is ambiguous. However, we are not emphasizing ambiguous grammars.

So (assuming a non-ambiguous grammar) the rightmost derivation in reverse can be obtained by constantly reducing the handle in the current string.

Given a grammar how do you find the handle (a handle if the grammar is ambiguous) of a string (which must be a right sentential form or there is no handle)?
Answer: Construct the (a if ambiguous, but we are not so interested in ambiguous grammars) rightmost derivation for the string and the handle is the last production you applied (so if you are doing rightmost derivations in reverse, the handle is the first production you would reduce by).

But how do you find the rightmost derivation?
Good question, we still have work to do.

Homework: 1, 2.

4.5.3: Shift-Reduce Parsing

We use two data structures for these parsers.

1. A stack of grammar symbols, terminals and nonterminals. This stack is drawn in examples as having its top on the right and bottom on the left. The items shifted (basically pushed, see below) onto the stack will be terminals, which are subsequently reduced to nonterminals. The bottom of the stack is marked with $ and initially the stack is empty (i.e., has just $).
2. An input buffer that (conceptually) holds the remainder of the input, i.e., the part that has yet to be shifted onto the stack. An endmarker $ is placed after the end of the input. Initially the input buffer contains the entire input followed by $. (In practice we use some more sophisticated buffering technique, as we saw in section 3.2 with buffers pairs that does not require having the entire input in memory at once.)

1. The parser can shift a symbol from the beginning of the input onto the TOS.
2. If the TOS is a handle, the parser can reduce it to its LHS.
3. If the parser reaches the state where the stack is $S (S is the start symbol) and the input is $, the parser terminates successfully.
4. The parser reaches an error state, where neither shifting nor reducing are possible.

A technical point, which explains the usage of a stack is that a handle is always at the TOS. See the book for a proof; the idea is to look at what rightmost derivations can do (specifically two consecutive productions) and then trace back what the parser will do since it does the reverse operations (reductions) in the reverse order.

We have not yet discussed how to decide whether to shift or reduce when both are possible. We have also not discussed which reduction to choose if multiple reductions are possible. These are crucial question for bottom up (shift-reduce) parsing and will be addressed.

Homework: 3.

4.6: Introduction to LR Parsing: Simple LR

I will have much more to say about SLR (simple LR) than the other LR schemes. The reason is that SLR is simpler to understand, but does capture the essence of shift-reduce, bottom-up parsing. The disadvantage of SLR is that there are LR grammars that are not SLR.

4.6.1: Why LR Parsers?

The text's presentation is somewhat controversial. Most commercial compilers use hand-written top-down parsers of the recursive-descent (LL not LR) variety. Since the grammars for these languages are not LL(1), the straightforward application of the techniques we have seen will not work. Instead the parsers actually look ahead further than one token, but only at those few places where the grammar is in fact not LL(1). Recall that (hand written) recursive descent compilers have a procedure for each nonterminal so we can customize as needed.

These compiler writers claim that they are able to produce much better error messages than can readily be obtained by going to LR (with its attendant requirement that a parser-generator be used since the parsers are too large to construct by hand). Note that compiler error messages are a very important user interface issue and that with recursive descent one can augment the procedure for a nonterminal with statements like
if (nextToken == X) then error(expected Y here)

Nonetheless, the claims made by the text are correct, namely.

1. LR parsers can be constructed to recognize nearly all programming-language constructs for which CFGs exist.
2. LR-parsing is the most general non-backtracking, shift-reduce method known, yet can be implemented relatively efficiently.
3. LR-parsing can detect a syntactic error as soon as possible.
4. LR grammars can describe more languages than LL grammars.

4.6.2: Items and the LR(0) Automaton

We now come to grips with the big question: How does a shift-reduce parser know when to shift and when to reduce? This will take a while to answer in a satisfactory manner. The unsatisfactory answer is that the parser has tables that say in each situation whether to shift or reduce (or announce error, or announce acceptance). To begin the path toward the answer, we need several definitions.

An item is a production with a marker saying how far the parser has gotten with this production. Formally,

Definition: An (LR(0)) item of a grammar is a production with a dot added somewhere to the RHS.

Note: A production with n symbols on the RHS, generates n+1 items.

Examples:

1. E → E + T generates 4 items.
   1. E → · E + T
   2. E → E · + T
   3. E → E + · T
   4. E → E + T ·
2. A → ε generates A → · as its only item.

The item E → E · + T signifies that the parser has just processed input that is derivable from E (i.e., reducible to E) and will look for input derivable from + T.

The item E → E + T · indicates that the parser has just seen the entire RHS and must consider reducing it to E.
Important: consider reducing does not mean reduce.

The parser groups certain items together into states. As we shall see, the items within a given state are treated similarly.

Our goal is to construct first the canonical LR(0) collection of states and then a DFA called the LR(0) automaton (technically not a DFA since it has no dead state).

To construct the canonical LR(0) collection formally and present the parsing algorithm in detail we shall

1. augment the grammar
2. define functions CLOSURE and GOTO

Augmenting the grammar is easy. We simply add a new start state S' and one production S'→S. The purpose is to detect success, which occurs when the parser is ready to reduce S to S'.

So our example grammar

    E → E + T | T
    T → T * F | F
    F → ( E ) | id
  

Interlude: The NFA Behind the Scenes

I hope the following interlude will prove helpful. When I was first preparing to present SLR a few years ago, I was struck by how much it looked like we were working with a DFA that came from some (unspecified and unmentioned) NFA. It seemed that by first doing the NFA, I could give some insight, especially since that is how we proceeded last chapter with lexers.

Since our current example would generate an NFA with many states and hence a big diagram, let's consider instead the following extremely simple grammar.

    E → E + T
    E → T
    T → id
  

    E' → E
    E  → E + T
    E  → T
    T  → id
  

Note that there are four kinds of edges.

1. Edges labeled with terminals. These correspond to shift actions, where the indicated terminal is shifted from the input to the stack.
2. Edges labeled with nonterminals. These will correspond to reduce actions when we construct the DFA. The stack is reduced by a production having the given nonterminal as LHS. Reduce actions do more as we shall see.
3. Edges labeled with ε. These are associated with the closure operation to be discussed and are the source of the nondeterminism (i.e., why the diagram is an NFA).
4. An edge labeled $. This edge, which can be thought of as shifting the endmarker, is used when we are reducing via the E'→E production and accepting the input.

If we are at the item E→E·+T (the dot indicating that we have seen an E and now need a +) and then shift a + from the input to the stack, we move to the item E→E+·T. If the dot is before a nonterminal, the parser needs a reduction with that nonterminal as the LHS.

Now we come to the idea of closure, which I illustrate in the diagram with the ε's. I hope this will help you understand the idea of closure, which like ε in regular expressions, leads to nondeterminism.

Look at the start state. The placement of the dot indicates that we next need to see an E. Since E is a nonterminal, we won't see it in the input, but will instead have to generate it via a production. Thus by looking for an E, we are also looking for any production that has E on the LHS. This is indicated by the two ε's leaving the top left box. Similarly, there are ε's leaving the other three boxes where the dot is immediately to the left of a nonterminal.

Remark: Perhaps instead of saying also looking for I should say really looking for.

As with regular expressions, we combine n-items connected by an ε arc into a d-item. The actual terminology used is that we combine these items into a set of items (later referred to as a state). For example all four items in the left column of the diagram above are combined into the state or item set labelled I₀ in the diagram on the right.

There is another combination that occurs. The top two n-items in the left column of the diagram above both have E transitions (outgoing arcs labeled E). Since we are considering these two n-items to be the same d-item and the arcs correspond to the same transition, the two targets (the top two n-items in the 2nd column of the diagram above) are combined. A d-item has all the outgoing arcs of the original n-items it contains. This is the way we converted an NFA into a DFA via the subset algorithm described in the previous chapter.

I₀, I₁, etc are called (LR(0)) item sets. The DFA containing these item sets as states and the state transitions described above is called the LR(0) automaton.

We start in the initial state with the symbols column empty and the input full. The $'s are just end markers. From state 0, called I0 in my diagram (following the book they are called I's since they are sets of items), we can only shift in the id (the nonterminals will appear in the symbols column). This brings us to I3 so we push a 3 onto the stack

In I3 we first notice that there is no outgoing arc labeled with a terminal; hence we cannot do a shift. However, we do see a completed production in the box (the dot is on the extreme right). Since the RHS consists solely of terminals, having the dot at the end means that we have seen (i.e., shifted) in the input the entire RHS of this production and are ready to perform a reduction. To reduce we pop the stack for each symbol in the RHS since we are replacing the RHS by the LHS. This time the RHS has one symbol so we pop the stack once and also remove one symbol from the symbols column. The stack corresponds to moves (i.e., state transtions) so we are undoing the move to 3 and we are temporarily in 0 again. But the production has a T on the LHS so we follow the T transition from 0 to 2, push T onto Symbols, and push 2 on.to the stack.

In I2 we again see no possible shift, but do see a completed production. This time the RHS contains a nonterminal so is not simply the result of shifting in symbols, but also reflects previous reductions. We again perform the indicated reduction, which takes us to I1

You might think that at I1 we could reduce using the completed bottom production, but that is wrong. This item (E'→E·) is special and can only be applied when we are at the end of the input string.

Thus the next two steps are shifts of + and id, sending us to 3 again, where, as before, we have no choice but to reduce the id to T and are in step 5 ready for the big one.

The production in 5 has three symbols on the RHS so we pop (back up) three times again temporarily landing in 0, but the RHS puts us in 1.

Perfect! We have just E as a symbol and the input is empty so we are ready to reduce by E'→E, which signifies acceptance.

Now we rejoin the book and say it more formally.

Actually there is more than mere formality coming. For the example above, we had no choices, but that was because the example was simple. We need more machinery to insure that we never have two or more possible moves to choose among. Specifically, we will need FOLLOW sets, the same ones we calculated for top-down parsing.

Closure of Item Sets

Say I is a set of items and one of these items is A→α·Bβ. This item represents the parser having seen α and records that the parser might soon see the remainder of the RHS. For that to happen the parser must first see a string derivable from B. Now consider any production starting with B, say B→γ. If the parser is to make progress on A→α·Bβ, it will need to be making progress on one such B→·γ. Hence we want to add all the latter productions to any state that contains the former. We formalize this into the notion of closure.

For any set of items I, CLOSURE(I) is formed as follows.

1. Initialize CLOSURE(I) = I
2. If A → α · B β is in CLOSURE(I) and B → γ is a production, then add B → · γ to the closure and repeat.

Example: Recall our main example

    E' → E
    E  → E + T | T
    T  → T * F | F
    F  → ( E ) | id
  

The Function GOTO

If X is a grammar symbol, then moving from A→α·Xβ to A→αX·β signifies that the parser has just processed (input derivable from) X. The parser was in the former position and (input derivable from) X was on the input; this caused the parser to go to the latter position. We (almost) indicate this by writing GOTO(A→α·Xβ,X) is A→αX·β. I said almost because GOTO is actually defined from item sets to item sets not from items to items.

Definition: If I is an item set and X is a grammar symbol, then GOTO(I,X) is the closure of the set of items A→αX·β where A→α·Xβ is in I.

The Canonical Collection of LR(0) Items—the LR(0) Automaton

I really believe this is very clear, but I understand that the formalism makes it seem confusing. Let me begin with the idea.

We augment the grammar and get this one new production; take its closure. That is the first element of the collection; call it I₀. Try GOTOing from I₀, i.e., for each grammar symbol, consider GOTO(I₀,X); each of these (almost) is another element of the collection. Now try GOTOing from each of these new elements of the collection, etc. Start with jane smith, add all her friends F, then add the friends of everyone in F, called FF, then add all the friends of everyone in FF, etc

The (almost) is because GOTO(I₀,X) could be empty so formally we construct the canonical collection of LR(0) items, C, as follows

1. Initialize C = CLOSURE({S' → S})
2. If I is in C, X is a grammar symbol, and GOTO(I,X)≠φ then add it to C and repeat.

This GOTO gives exactly the arcs in the DFA I constructed earlier. The formal treatment does not include the NFA, but works with the DFA from the beginning.

Definition: The above collection of item sets (so this is a set of sets) is called the canonical LR(0) collection and the DFA having this collection as nodes and the GOTO function as arcs is called the LR(0) automaton.

Homework: Construct the LR(0) automaton for the following grammar (which produces simple postfix expressions).
  S → S S + | S S * | a
Don't forget to augment the grammar.

The DFA for our Main Example

Our main example

    E' → E
    E  → E + T | T
    T  → T * F | F
    F  → ( E ) | id
  

Start constructing the diagram on the board:
Begin with {E' → ·E}, take the closure, and then keep applying GOTO.

Start Lecture #7

Remark: Lab 3 assigned. Part 1 (no programming) due in one week; the remainder due a week later.

Use of the LR(0) Automaton

A state of the automaton is an item set as described previously. The transition function is GOTO. If during a parse we are up to item set I_j (often called state s_j or simply state j) and the next input symbol is b (it of course must be a terminal), then the parser shifts in b if the state j has an outgoing transition labeled b. If there is no such transition, then the parser performs a reduction; choosing which reduction to use is determined by the items in I_j and the FOLLOW sets. (It is also possible that the parser will now accept the input string or announce that the input string is not in the language).

4.6.3: The LR-Parsing Algorithm

The LR-parsing algorithm must decide when to shift and when to reduce (and in the latter case, by which production). It does this by consulting two tables, ACTION and GOTO. The basic algorithm is the same for all LR parsers, what changes are the tables ACTION and GOTO.

The LR Parsing Tables

We have already seen GOTO (for SLR).

Technical point that may, and probably should, be ignored: our GOTO was defined on pairs [item-set,grammar-symbol]. The new GOTO is defined on pairs [state,nonterminal]. A state is simply an item set (so nothing is new here). We will not use the new GOTO on terminals so we just define it on nonterminals.

Given a state i and a terminal a (or the endmarker), ACTION[i,a] can be

1. Shift j. The terminal a is shifted on to the stack and the parser enters state j.
2. Reduce A → α. The parser reduces α on the TOS to A.
3. Accept.
4. Error

So ACTION is the key to deciding shift vs. reduce. We will soon see how this table is computed for SLR.

Since ACTION is defined on [state,terminal] pairs and GOTO is defined on [state,nonterminal] pairs, we can combine these tables into one defined on [state,grammar-symbol] pairs.

LR-Parser Configurations (formalism)

This formalism is useful for stating the actions of the parser precisely, but I believe the parser can be explained without this formalism. The essential idea of the formalism is that the entire state of the parser can be represented by the vector of states on the stack and input symbols not yet processed.

Behavior of the LR Parser

The parser consults the combined ACTION-GOTO table for its current state (TOS) and next input symbol, formally this is ACTION[s_m,a_i], and proceeds based on the value in the table. If the action is a shift, the next state is clear from the DFA We have done this informally just above; here we use the formal treatment).

1. Shift s. The input symbol a is pushed and s becomes the new state. The new configuration is
   (s₀...s_ms,a_i+1...a_n)
2. Reduce A → α. Let r be the number of symbols in the RHS of the production. The parser pops r items off the stack (backing up r states) and enters the state GOTO(s_m-r,A). That is after backing up it goes where A says to go. A real parser would now probably do something, e.g., build a tree node or perform a semantic action. Although we know about this from the chapter 2 overview, we don't officially know about it here. So for now think of it simply printing the production the parser reduced by.
3. Accept.
4. Error.

4.6.4 Constructing SLR-Parsing Tables

The missing piece of the puzzle is finally revealed.

A Simple (but fake) Example

Before defining the ACTION and GOTO tables precisely, I want to do it informally via the simple example on the right. This is a FAKE example, the B'→B· item would be in I₀ and would have only B→·a+b with it.

IMPORTANT: To construct the table, you do need something not in the diagram, namely the FOLLOW sets from top-down parsing.

For convenience number the productions of the grammar to make them easy to reference and assume that the production B → a+b is numbered 2. Also assume FOLLOW(B)={b} and all other follow sets are empty. Again, I am not claiming that there is a grammar with this diagram and these FOLLOW sets.

Show how to calculate every entry in the table using the diagram and the FOLLOW sets. Then consider input a+b and

The action table is defined with states (item sets) as rows and terminals and the $ endmarker as columns. GOTO has the same rows, but has nonterminals as columns. So we construct a combined ACTION-GOTO table, with states as rows and grammar symbols (terminals + nonterminals) plus $ as columns.

1. Each arc in the diagram labeled with a terminal indicates a shift. In the entry with row the state at the tail of the arc and column the labeling terminal place sn, where n is the state at the head of the arc. This indicates that if we are in the given state and the input is the given terminal, we shift to new state n.
2. Each arc in the diagram labeled with a nonterminal informs us what state to enter if we reduce. In the entry with row the state at the tail of the arc and column the labeling nonterminal place n, where n is the state at the head of the arc.
3. Each completed item (dot at the extreme right) indicates a possible reduction. In each entry with row the state containing the completed item and column a terminal in the FOLLOW set of the LHS of the production corresponding to this item, place rn, where n is the number of the production. (In particular, at entry [13,b] place an r2.).
4. In the entry with row (i.e., state) containing S'→S and column $, place accept.
5. If any entry is labelled twice (i.e., a conflict) the grammar is not SLR(1).
6. Any unlabeled entry corresponds to an input error. If the parser accesses this entry, the input sentence is not in the language generated by the grammar.

The Table Construction Algorithm

The diagrams above are constructed for our use; they are not used in constructing a bottom-up parser. Here is the real algorithm, which uses just the augmented grammar (i.e., after adding S' → S) and the FOLLOW sets.

1. Construct {I₀,...,I_n} the LR(0) items.
2. The parsing actions for state i.
   1. If A→α·bβ is in I_i for b a terminal, then ACTION[i,b]=shift j, where GOTO(I_i,b)=I_j.
   2. If A→α· is in I_i, for A≠S', then, for all b in FOLLOW(A), ACTION[i,b]=reduce A→α.
   3. If S'→S· is in I_i, then ACTION[I,$]=accept.
   4. If any conflicts occurred, the grammar is not SLR(1).
3. If GOTO(I_i,A)=I_j, for a nonterminal A, then GOTO[i,A]=j.
4. All entries not yet defined are error.
5. The initial state is the one containing S'→·S.

The Main Example

Our main example, pictured on the right, gives the table shown on the left. The productions and FOLLOW sets are shown as well (the FIRST sets are not used directly, but are needed to calculate FOLLOW). The table entry s5 abbreviates shift and go to state 5. The table entry r2 abbreviates reduce by production number 2, where we have numbered the productions as follows.

The shift actions can be read directly off the DFA. For example I1 with a + goes to I6, I6 with an id goes to I5, and I9 with a * goes to I7.

The reduce actions require FOLLOW, which for this simple grammar is fairly easy to calculate.

Consider I₅={F→id·}. Since the dot is at the end, we are ready to reduce, but we must check if the next symbol can follow the F we are reducing to. Since FOLLOW(F)={+,*,),$}, in row 5 (for I5) we put r6 (for reduce by production 6) in the columns for +, *, ), and $.

The GOTO columns can also be read directly off the DFA. Since there is an E-transition (arc labeled E) from I₀ to I₁, the column labeled E in row 0 contains a 1.

Since the column labeled + is blank for row 7, we see that it would be an error if we arrived in state 7 when the next input character is +.

Finally, if we are in state 1 when the input is exhausted ($ is the next input character), then we have a successfully parsed the input.

Homework: 2 (you already constructed the LR(0) automaton for this example in the previous homework), 3, 4 (this problem refers to 4.2.2(a-g); only use 4.2.2(a-c).

Example: What about ε-productions? Let's do

    A → B D
    B → b B | ε
    D → d
  

Reducing by the ε-production actually adds a state to the stack (it pops ZERO states since there are zero symbols on RHS and pushes one).

Homework: Do the following extension

    A → B D
    B → b B | ε
    D → d D | ε
  

4.6.5: Viable Prefixes

4.7: More Powerful LR Parsers

We consider very briefly two alternatives to SLR, canonical-LR or LR, and lookahead-LR or LALR.

4.7.1: Canonical LR(1) Items

SLR used the LR(0) items, that is the items used were productions with an embedded dot, but contained no other (lookahead) information. The LR(1) items contain the same productions with embedded dots, but add a second component, which is a terminal (or $). This second component becomes important only when the dot is at the extreme right (indicating that a reduction can be made if the input symbol is in the appropriate FOLLOW set). For LR(1) we do that reduction only if the input symbol is exactly the second component of the item. This finer control of when to perform reductions, enables the parsing of a larger class of languages.

4.7.4: Constructing LALR Parsing Tables

For LALR we merge various LR(1) item sets together, obtaining nearly the LR(0) item sets we used in SLR. LR(1) items have two components, the first, called the core, is a production with a dot; the second a terminal. For LALR we merge all the item sets that have the same cores by combining the 2nd components (thus permitting reductions when any of these terminals is the next input symbol). Thus we obtain the same number of states (item sets) as in SLR since only the cores distinguish item sets.

Unlike SLR, we limit reductions to occurring only for certain specified input symbols. LR(1) gives finer control; it is possible for the LALR merger to have reduce-reduce conflicts when the LR(1) items on which it is based is conflict free.

Although these conflicts are possible, they are rare and the size reduction from LR(1) to LALR is quite large. LALR is the current method of choice for bottom-up, shift-reduce parsing.

4.9: Parser Generators

4.9.1: The Parser Generator Yacc

The tool corresponding to Lex for parsing is yacc, which (at least originally) stood for yet another compiler compiler. This name is cute but somewhat misleading since yacc (like the previous compiler compilers) does not produce a compiler, just a parser.

The structure of the yacc user input is similar to that for lex, but instead of regular definitions, one includes productions with semantic actions.

There are ways to specify associativity and precedence of operators. It is not done with multiple grammar symbols as in a pure parser, but more like declarations.

Use of Yacc requires a serious session with its manual.

Start Lecture #8

Remark: Midterm grades are based on labs 1 and 2. Note that this is only ~25% of the final grade.

Chapter 5: Syntax-Directed Translation

Homework: Read Chapter 5.

Again we are redoing, more formally and completely, things we briefly discussed when breezing over chapter 2.

Recall that a syntax-directed definition (SDD) adds semantic rules to the productions of a grammar. For example to the production T → T / F we might add the rule

    T.code = T1.code || F.code || '/'
  

Rather than constantly copying ever larger strings to finally output at the root of the tree after a depth first traversal, we can perform the output incrementally by embedding semantic actions within the productions themselves. The above example becomes

    T → T1 / F { print '/' }
  

Since we are generating postfix, the action comes at the end (after we have generated the subtrees for T₁ and F, and hence performed their actions). In general the actions occur within the production, not necessarily after the last symbol.

For SDD's we conceptually need to have the entire tree available after the parse so that we can run the postorder traversal. (It is postorder since we have a simple S-attributed SDD; we will traverse the parse tree in other orders when the SDD is not S-attributed, and will see situations when no traversal order is possible.)

In some situations semantic actions can be performed during the parse, without saving the tree.

5.1: Syntax-Directed Definitions (SDDs)

Formally, attributes are values (of any type) that are associated with grammar symbols. Write X.a for the attribute a of symbol X. You can think of attributes as fields in a record/struct/object.

Semantic rules (rules for short) are associated with productions and give formulas to calculate attributes of nonterminals in the production.

5.1.1: Inherited and Synthesized Attributes

Terminals can have synthesized attributes (not yet officially defined); these are given to it by the lexer (not the parser). For example the token id might well have the attribute lexeme, where id.lexeme is the lexeme that the lexer converted into this instance of id. There are no rules in an SDD giving attribute values to terminals. Terminals do not have inherited attributes (to be defined shortly).

A nonterminal B can have both inherited and synthesized attributes. The difference is how they are computed. In either case the computation is specified by a rule associated with the production at a node N of the parse tree.

Definition: A synthesized attribute of a nonterminal B is defined at a node N where B is the LHS of the production associated with N. The attribute can depend on only (synthesized or inherited) attribute values at the children of N (the RHS of the production) and on inherited attribute values at N itself.

The arithmetic division example above was synthesized.

Example: The SDD at the right gives a left-recursive grammar for expressions with an extra nonterminal L added as the start symbol. The terminal num is given a value by the lexer, which corresponds to the value stored in the numbers table for lab 2.

Draw the parse tree for 7+6/3 on the board and verify that L.val is 9, the value of the expression.

This example uses only synthesized attributes.

Definition: An SDD with only synthesized attributes is called S-attributed.

For these SDDs all attributes depend only on attribute values at the children.
Why?
Answer: The other possibility (depending on values at the node itself) involves inherited attributes, which are absent in an S-attributed SDD.

Thus, for an S-attributed SDD, all the rules can be evaluated by a single bottom-up (i.e., postorder) traversal of the annotated parse tree.

Inherited attributes are more complicated since the node N of the parse tree with which the attribute is associated (and which is also the natural node to store the value) does not contain the production with the corresponding semantic rule.

Definition: An inherited attribute of a nonterminal B at node N, where B is the LHS of the production, is defined by a semantic rule of the production at the parent of N (where B occurs in the RHS of the production). The value depends only on attributes at N, N's siblings, and N's parent.

Note that when viewed from the parent node P (the site of the semantic rule), the inherited attribute depends on values at P and at P's children (the same as for synthesized attributes). However, and this is crucial, the nonterminal B is the LHS of a child of P and hence the attribute is naturally associated with that child. It is normally stored there and is shown there in the diagrams below.

We will soon see examples with inherited attributes.

Definition: Often the attributes are just evaluations without side effects. In such cases we call the SDD an attribute grammar.

5.1.2: Evaluating an SDD at the Nodes of a Parse Tree

If we are given an SDD and a parse tree for a given sentence, we would like to evaluate the annotations at every node. Since, for synthesized annotations parents can depend on children, and for inherited annotations children can depend on parents, there is no guarantee that one can in fact find an order of evaluation. The simplest counterexample is the single production A→B with synthesized attribute A.syn, inherited attribute B.inh, and rules A.syn=B.inh and B.inh=A.syn+1. This means to evaluate A.syn at the parent node we need B.inh at the child and vice versa. Even worse it is very hard to tell, in general, if every sentence has a successful evaluation order.

All this not withstanding we will not have great difficulty because we will not be considering the general case.

Annotated Parse Trees

Recall that a parse tree has leaves that are terminals and internal nodes that are nonterminals.

Definition: A parse tree decorated with attributes, is called an annotated parse tree. It is constructed as follows.

Each internal node corresponds to a production. The node is labeled with the LHS of the production. If there are no attributes for the LHS in this production, we leave the node as it was (I don't believe this is a common occurrence). If there are k attributes for the LHS, we replace the LHS in the parse tree by k equations. The LHS of the equation is the attribute and the right hand side is its value.

Note that the annotated parse tree contains all the information of the original parse tree since we (either leave the node alone—if the LHS had no attributes or) replace the nonterminal A labeling the LHS with a series of equations A.attr=value.

We computed the values to put in this tree for 7+6/3 and on the right is (7-6).

Homework: 1

Why Have Inherited Attributes?

Inherited attributed definitely make the situation more complicated. For a simple example, recall the circular dependency above involving A.syn and B.inh. But we do need them.

Consider the following left-recursive grammar for multiplication of numbers, and the parse tree on the right for 3*5*4.

    T → T * F
    T → F
    F → num
  

It is easy to see how the values can be propagated up the tree and the expression evaluated.

When doing top-down parsing, however, we need to avoid left recursion. Consider the grammar below, which is the result of removing the left recursion. Again its parse tree is shown on the right. Try not to look at the semantic rules for the moment.

Where on the tree should we do the multiplication 3*5 since there is no node that has 3 and * and 5 as children?

The second production is the one with the *, so that is the natural candidate for the multiplication site. Make sure you see that this production (for the * in 3*5) is associated with the blue-highlighted node in the parse tree. The right operand (5) can be obtained from the F that is the middle child of this T'. F gets the value from its child, the number itself; this is an example of the simple synthesized case we have already seen, F.val=num.lexval (see the last semantic rule in the table).

But where is the left operand? It is located at the sibling of T' in the parse tree, i.e., at the F immediately to T's left (we shall see the significance that the sibling is to the left; it is not significant that it is immediately to the left). This F is not mentioned in the production associated with the T' node we are examining.
So, how does T' get F.val from its sibling?
Answer: The common parent, in this case T, can get the value from F and then our node can inherit the value from its parent.
Bingo! ... an inherited attribute. This can be accomplished by having the following two rules at the node T.
    T.tmp   = F.val (synthesized)
    T'.lval = T.tmp (inherited)

Since we have no other use for T.tmp, we combine the above two rules into the first rule in the table.

Now lets look at the second multiplication (3*5)*4, where the parent of T' is another T'. (This is the normal case. When there are n multiplies, n-1 have T' as parent and only one has T).

The pink-highlighted T' is the site for the multiplication. However, it needs as left operand, the product 3*5 that its parent can calculate. So we have the parent (another T' node, the blue one in this case) calculate the product and store it as an attribute of its right child namely the pink T'. That is the first rule for T' in the table.

We have now explained the first, third, and last semantic rules. These are enough to calculate the answer. Indeed, if we trace it through, 60 does get evaluated and stored in the bottom right T', the one associated with the ε-production. Our remaining goal is to get the value up to the root where it represents the evaluation of this term T and can be combined with other terms to get the value of a larger expression.

Going up is easy, just synthesize. I named the attribute tval, for term-value. It is generated at the ε-production from the lval attribute and propagated back up. At the T node it is called simply val. At the right we see the annotated parse tree for this input.

Homework: Extend this SDD to handle the left-recursive, more complete expression evaluator given earlier in this section. Don't forget to eliminate the left recursion first.

It clearly requires some care to write the annotations.

Another question is how does the system figure out the evaluation order, assuming one exists? That is the subject of the next section.

Remark: Consider the identifier table. The lexer creates it initially, but as the compiler performs semantic analysis and discovers more information about various identifiers, e.g., type and visibility information, the table is updated. One could think of this as some sort of inherited/synthesized attribute pair that during each phase of analysis is pushed down and back up the tree. However, it is not implemented this way; the table is made a global data structure that is simply updated. The compiler writer must ensure manually that the updates are performed in an order respecting any dependences.

5.2: Evaluation Orders for SDD's

5.2.1: Dependency Graphs

The diagram on the right illustrates a great deal. The black dotted lines comprise the parse tree for the multiplication grammar just studied when applied to a single multiplication, e.g. 3*5.

Each synthesized attribute is shown in green and is written to the right of the grammar symbol at the node where it is defined.

Each inherited attribute is shown in red and is written to the left of its grammar symbol where its value is assigned. Note that this value is defined by semantic rule at the parent of this node.

Each green arrow points to the synthesized attribute calculated from the attribute at the tail of the arrow. These arrows either go up the tree one level or stay at a node. That is because a synthesized attribute can depend only on the node where it is defined and that node's children. The computation of the attribute is associated with the production at the node at its arrowhead. In this example, each synthesized attribute depends on only one other attribute, but that is not required.

Each red arrow points to the inherited attribute calculated from the attribute at the tail. Note that, at the lower right T' node, two red arrows point to the same attribute. This indicates that the common attribute at the arrowheads, depends on both attributes at the tails. According to the rules for inherited attributes, these arrows either go down the tree one level, go from a node to a sibling, or stay within a node. The computation of the attribute is associated with the production at the parent of the node at the arrowhead.

5.2.2: Ordering the Evaluation of Attributes

The graph just drawn is called the dependency graph. In addition to being generally useful in recording the relations between attributes, it shows the evaluation order(s) that can be used. Since the attribute at the head of an arrow depends on the on the one at the tail, we must evaluate the head attribute after evaluating the tail attribute.

Thus what we need is to find an evaluation order respecting the arrows. This is called a topological sort. The rule is that the needed ordering can be found if and only if there are no (directed) cycles. The algorithm is simple.

1. Choose a node having no incoming edges
2. Delete the node and all outgoing edges.
3. Repeat

If the algorithm succeeds in deleting all the nodes, then the deletion order is a suitable evaluation order and there were no directed cycles.

The topological sort algorithm is non-deterministic (Choose a node) and hence there can be many topological sort orders.

Homework: 1.

5.2.3: S-Attributed Definitions

Given an SDD and a parse tree, performing a topological sort constructs a suitable evaluation order or announces that no such order exists.

However, it is very difficult to determine, given just an SDD, whether no parse trees have cycles in their dependency graphs. That is, it is very difficult to determine if there are suitable evaluation orders for all parse trees. Fortunately, there are classes of SDDs for which a suitable evaluation order is guaranteed, and which are adequate for our needs.

As mentioned above an SDD is S-attributed if every attribute is synthesized. For these SDDs all attributes are calculated from attribute values at the children since the other possibility, the tail attribute is at the same node, is impossible since the tail attribute must be inherited for such arrows. Thus no cycles are possible and the attributes can be evaluated by a postorder traversal of the parse tree.

5.2.4 L-Attributed Definitions

Unfortunately, it is hard to live without inherited attributes. For example, we need them for top-down parsing of expressions. Fortunately, our needs can be met by a class of SDDs called L-Attributed definitions for which we can easily find an evaluation order.

Case 2c must be handled specially whenever it occurs. We will try to avoid it.

The top picture to the right illustrates the other cases and suggests why there cannot be any cycles.

The picture immediately to the right corresponds to a fictitious R-attributed definition. One reason L-attributed definitions are favored over R, is the left to right ordering in English. See the example below on type declarations and also consider the grammars that result from eliminating left recursion.

We shall see that the key property of L-attributed SDDs is that they can be evaluated with two passes over the tree (an euler-tour order) in which we evaluate the inherited attributes as we go down the tree and the synthesized attributes as we go up. The restrictions L-attributed SDDs place on the inherited attributes are just enough to guarantee that when we go down we have all the values needed for the inherited attributes of the child.

Evaluating L-Attributed Definitions

The table on the right shows a very simple grammar with fairly general, L-attributed semantic rules attached. Compare the dependencies with the general case shown in the (red-green) picture of L-attributed SDDs above.

The picture below the table shows the parse tree for the grammar in the table. The triangles below B and C represent the parse tree for X and Y. The dotted and numbered arrow in the picture illustrates the evaluation order for the attributes; it will be discussed shortly.

The rules for calculating A.syn, B.inh, and C.inh are shown in the table. The attribute A.inh would have been set by the parent of A in the tree; the semantic rule generating A.inh would be given with the production at the parent. The attributes X.syn and Y.syn are calculated at the children of B and C respectively. X.syn can depend of X.inh and on values in the triangle below X; similarly for Y.syn.

The picture to the right shows that there is an evaluation order for L-attributed definitions (again assuming no case 2c). We just need to follow the (Euler-tour) arrow and stop at all the numbered points. As in the pictures above, red signifies inherited attributes and green synthetic. Specifically, the evaluations at the numbered stops are

1. A is invoked (viewing the traversal as a program) and is passed its inherited attributes (A.inh in our case, but of course there could be several such attributes), which have been evaluated at its parent.
2. B is invoked by A and is given B.inh, which A has calculated. In programming terms: A executes
       call B(B.inh)
   where the argument has been evaluated by A. This argument can depend on A.inh since the parent of A has given A this value.
3. B calls its first child (in our example X is the only child) and passes to the child the child's inherited attributes. In programming terms: B executes
       call X(X.inh)
4. The child returns, passing back to B the synthesized attributes of the child. In programming terms: X executes
       return X.syn
   In reality there could be more synthesized attributes, there could be more children, the children could have children, etc.
5. B returns to A passing back B.syn, which can depend on B.inh (given to B by A in step 2) and X.syn (given to B by X in the previous step).
6. A calls C giving C its inherited attributes, which can depend on A.inh (given to A, by A's parent), B.inh (previously calculated by A in step 2), and B.syn (given to A by B in step 5).
7. C calls its first child, just as B did.
8. The child returns to C, just as B's child returned to B.
9. C returns to A passing back C.syn, just as B did.
10. A returns to its parent passing back A.syn, which can depend on A.inh (given to A by its parent in step 1), B.inh calculated by A in step 2, B.syn (given to A by B in step 5), C.inh (calculated by A in step 6), and C.syn (given to A by C in step 9).

More formally, do a depth first traversal of the tree and evaluate inherited attributes on the way down and synthetic attributes on the way up. This corresponds to a an Euler-tour traversal. It also corresponds to a call graph of a program where actions are taken at each call and each return

The first time you visit a node (on the way down), evaluate its inherited attributes (in programming terms, the parent evaluates the inherited attributes of the children and passes them as arguments to the call). The second time you visit a node (on the way back up), you evaluate the synthesized attributes (in programming terms the child returns the value to the parent).

The key point is that all attributes needed will have already been evaluated. Consider the rightmost child of the root in the diagram on the right.

1. Inherited attributes (which are evaluated on the first, i.e., downward, pass): An inherited attribute depends only on inherited attributes from the parent and on (inherited or synthesized) attributes from left siblings.
   • The parent will have already been evaluated on the downward pass before the current child so the parent's inherited attributes will have already been evaluated.
   • The left children have already had both passes done so all their attributes will have been evaluated.
2. Synthesized attributes (which are evaluated on the second, i.e., upward pass): A synthesized attribute depends only on (inherited or synthesized) attributes of its children and on its own inherited attributes.
   • The children have already had both passes so all their attributes have been evaluated.
   • The node itself has already had its first (downward) pass so has had its inherited attributes evaluated.

Homework: 3(a-c).

5.2.5: Semantic Rules with Controlled Side Effects

When we have side effects such as printing or adding an entry to a table we must ensure that we have not added a constraint to the evaluation order that causes a cycle.

For example, the left-recursive SDD shown in the table on the right propagates type information from a declaration to entries in an identifier table.

The function addType adds the type information in the second argument to the identifier table entry specified in the first argument. Note that the side effect, adding the type info to the table, does not affect the evaluation order.

Draw the dependency graph on the board. Note that the terminal ID has an attribute called entry (given by the lexer) that gives its entry in the identifier table. The nonterminal L can be considered to have (in addition to L.type) a dummy synthesized attribute, say AddType, that is a place holder for the addType() routine. AddType depends on the arguments of addType(). Since the first argument is from a child, and the second is an inherited attribute of this node, we have legal dependences for a synthesized attribute.

Thus we have an L-attributed definition.

Homework: For the SDD above, give the annotated parse tree for

    INT a,b,c
  

5.3: Applications of Syntax-Directed Translations

5.3.1: Construction of Syntax Trees

Recall that a syntax tree (technically an abstract syntax tree) contains just the essential nodes. For example, 7+3*5 would have one + node, one *, and the three numbers. Lets see how to construct the syntax tree from an SDD.

Assume we have two functions Leaf(op,val) and Node(op,c1,...,cn), that create leaves and interior nodes respectively of the syntax tree. Leaf is called for terminals. Op is the label of the node (op for operation) and val is the lexical value of the token. Node is called for nonterminals and the ci's are pointers to the children.

The upper table on the right shows a left-recursive grammar that is S-attributed (so all attributes are synthesized).

Try this for x-2+y and see that we get the syntax tree.

When we eliminate the left recursion, we get the lower table on the right. It is a good illustration of dependencies. Follow it through and see that you get the same syntax tree as for the left-recursive version.

Remarks:

1. You probably did/are-doing/will-do some variant of new Node and new Leaf for lab 3. When processing a production
   1. Create a parse tree node for the LHS.
   2. Call subroutines for RHS symbols and connect the resulting nodes to the node created in i.
   3. Return a reference to the new node so the parent can hook it into the parse tree.
2. It is the lack of a call to new in the third and fourth productions (of the first, left recursive, grammar) that causes the (abstract) syntax tree to be produced rather than the parse (concrete syntax) tree.
3. Production compilers do not produce a parse trees; rather they produce syntax trees. The syntax tree is smaller, and hence more (space and time) efficient for subsequent passes that walk the tree. The parse tree is (I believe) slightly easier to construct as you don't have to decide which nodes to produce; you simply produce them all.
4. We could, but will not, have a lab 3.5 that takes in the parse tree generated by lab 3 and produces an abstract syntax tree using an SDD as illustrated by the table above.

5.3.2: The Structure of a Type

This course emphasizes top-down parsing (at least for the labs) and hence we must eliminate left recursion. The resulting grammars often need inherited attributes, since operations and operands are in different productions. But sometimes the language itself demands inherited attributes. Consider two ways to declare a 3x4, two-dimensional array. (Strictly speaking, these are one-dimensional arrays of one-dimensional arrays.)

    array [3] of array [4] of int    and     int[3][4]
  

Assume that we want to produce a tree structure like the one the right for either of the array declarations on the left. The tree structure is generated by calling a function array(num,type). Our job is to create an SDD so that the function gets called with the correct arguments.

For the first language representation of arrays (found in Ada and in lab 3), it is easy to generate an S-attributed (non-left-recursive) grammar based on
A → ARRAY [ NUM ] OF A | INT | REAL
This is shown in the upper table on the right.

On the board draw the parse tree and see that simple synthesized attributes above suffice.

For the second language representation of arrays (the C-style), we need some smarts (and some inherited attributes) to move the int all the way to the right. Fortunately, the result, shown in the table on the right, is L-attributed and therefore all is well.

Note that, instead of a third column stating whether the attribute is synthesized or inherited, I have adopted the convention of drawing the inherited attribute definitions with a pink background.

Also note that this is not necessary. That is, one can look at a production and the associated semantic rule and determine if the attribute is inherited or synthesized.
How is this done?
Answer: If the attribute being defined (i.e., the one on the LHS of the semantic rule) is associated with the nonterminal on the LHS of the production, the attribute is synthesized. If the attribute being defined is associated with a nonterminal on the RHS of the production, the attribute is inherited.

Homework: 1.

Start Lecture #9

5.5: Implementing L-Attributed SDD's

A good summary of the available techniques.

1. Build the parse tree and annotate. Works as long as no cycles are present (guaranteed by L- or S-attributed). This is the method we are using in the labs.
2. Build the parse tree, add actions, and execute the actions in preorder. Works for any L-attributed definition. Can add actions based on the semantic rules of the SDD. (Since actions are leaves of the tree, I don't see why preorder is relevant).
3. Translate During Recursive Descent Parsing. This is basically combining labs 3 and 4. See below. Production compilers often use more complicated semantic rules than ours and make multiple passes up and down the tree, which renders this optimization impossible.
4. Generate Code on the Fly. Also uses recursive descent, but is restrictive.
5. Implement an SDT during LL-parsing.
6. Implement an SDT during LR-parsing of an LL Language.

5.5.1: Translation During Recursive-Descent Parsing

Recall that in fully-general, recursive-descent parsing one normally writes one procedure for each nonterminal.

Assume the SDD is L-attributed.

We are interested in using predictive parsing for LL(1) grammars In this case we have a procedure P for each production, which is implemented as follows.

1. P is passed the inherited attributes of its LHS nonterminal.
2. P defines a variable for each attribute it will calculate. These are the inherited attributes of the nonterminals in the RHS and the synthesized attributes of the nonterminal constituting the LHS.
3. P processes in left-to-right order each nonterminal B appearing in the body of the production as follows.
   1. Calculate the inherited attributes of B. Since the SDD is L-attributed, these attributes do not depend on attributes for nonterminals to the right of B (which have not yet been processed).
   2. Call the procedure Q associated with this use of the nonterminal B. Since the grammar is LL(1) and we are using predictive parsing, we know what production with LHS B will be used here.
   3. The procedure Q returns to P the synthesized attributes of B, which may be needed by nonterminals to the right of B.
4. P calculates the synthesized attributes of the nonterminal constituting its LHS.
5. P returns these synthesized attributes to its caller (which is the procedure associated with its parent).

5.A: What Is This All Used For?

Assume we have the parse tree on the right as produced, for example, by your lab3. (The numbers should really be a token, e.g. NUM, not the lexemes shown.) You now want to write the semantics analyzer, or intermediate code generator, or lab 4. In any of these cases you have semantic rules or actions (for lab4 it will be semantic rules) that need to be performed. Assume the SDD or SDT is L-attributed (that is my job for lab4), so we don't have to worry about dependence loops.

You start to write
      analyze-i.e.-traverse (tree-node)
which will be initially called with tree-node=the-root. The procedure will perform an Euler-tour traversal of the parse tree and during its visits of the nodes, it will evaluate the relevant semantic rules.

The visit() procedure is basically a big switch statement where the cases correspond to evaluating the semantic rules for the different productions in the grammar. The tree-node is the LHS of the production and the children are the RHS.

By first switching on the tree-node and then inspecting enough of the children, visit() can tell which production the tree-node corresponds to and which semantic rules to apply.

As described in 5.5.1 above, visit() has received as parameters (in addition to tree-node), the inherited attributes of the node. The traversal calls itself recursively, with the tree-node argument set to the leftmost child, then calls again using the next child, etc. Each time, the child is passed the inherited attributes.

When each child returns, it passes back its synthesized attributes.

After the last child returns, the parent returns, passing back the synthesized attributes that were calculated.

A programming point is that, since tree-node can be any node, traverse() and visit() must each be prepared to accept as parameters any inherited attribute that any nonterminal can have.

5.A.A: Variations

1. Instead of a giant switch, you could have separate routines for each nonterminal and just switch on the productions having this nonterminal as LHS.

   In this case each routine need be prepared to accept as parameters only all the inherited attributes that its nonterminal can have.

2. You could have separate routines for each production and thus each routine has as parameters exactly the inherited attributes that this production receives.

   To do this requires knowing which visit() procedure to call for each nonterminal (child node of the tree). For example, assume you are processing the production
       B → C D
   You need to know which production to call for C (remember that C can be the LHS of many different productions).

   • You could store the (number of the) production in the tree node.
   • You could peek further down the tree to determine the production for each of your children (i.e., you look at grandchildren).
   • You could try all productions for a given nonterminal, telling the child what production you think it is and having the child return failure when you are wrong. This seems too much work to me.
   • etc
   
   
3. If you like actions instead of rules, perform the actions where indicated in the SDT.
   
   
4. Global variable can be used (with care) instead of parameters.
   
   
5. As illustrated earlier, you can call routines instead of setting an attribute (see addType in 5.2.5).

Chapter 6: Intermediate-Code Generation

Homework: Read Chapter 6.

6.1: Variants of Syntax Trees

6.1.1: Directed Acyclic Graphs for Expressions

The difference between a syntax DAG and a syntax tree is that the former can have undirected cycles. DAGs are useful where there are multiple, identical portions in a given input. The common case of this is for expressions where there often are common subexpressions. For example in the expression
      X + a + b + c - X + ( a + b + c )
each individual variable is a common subexpression. But a+b+c is not since the first occurrence has the X already added. This is a real difference when one considers the possibility of overflow or of loss of precision. The easy case is
      x + y * z * w - ( q + y * z * w )
where y*z*w is a common subexpression.

It is easy to find such common subexpressions. The constructor Node() above checks if an identical node exists before creating a new one. So Node ('/',left,right) first checks if there is a node with op='/' and children left and right. If so, a reference to that node is returned; if not, a new node is created as before.

Homework: 1.

6.1.2: The Value-Number Method for Constructing DAGS

Often one stores the tree or DAG in an array, one entry per node. Then the array index, rather than a pointer, is used to reference a node. This index is called the node's value-number and the triple
    <op, value-number of left, value-number of right>
is called the signature of the node. When Node(op,left,right) needs to determine if an identical node exists, it simply searches the table for an entry with the required signature.

Searching an unordered array is slow; there are many better data structures to use. Hash tables are a good choice.

Homework: 2.

6.2: Three-Address Code

We will use three-address code, i.e., instructions of the form op a,b,c, where op is a primitive operator. For example

    lshift a,b,4   // left shift b by 4 and place result in a
    add    a,b,c   // a = b + c
    a = b + c      // alternate (more natural) representation of above
  

If we are starting with an expression DAG (or syntax tree if less aggressive), then transforming into 3-address code is just a topological sort and an assignment of a 3-address operation with a new name for the result to each interior node (the leaves already have names and values).

A key point is that nodes in an expression dag (or tree) have at most 2 children so three-address code is easy. As we produce three-address code for various constructs, we may have to generate several instructions to translate one construct.

For example, (B+A)*(Y-(B+A)) produces the DAG on the right, which yields the following 3-address code.

    t1 = B + A
    t2 = Y - t1
    t3 = t1 * t2
  

Notes

1. We have not yet learned how to have the compiler generate the instructions given above.
2. The labs will use parse trees not syntax dags.

6.2.1: Addresses and Instructions

We use the terminology 3-address code since instructions in our intermediate-language consist of one elementary operation with three operands, each of which is often an address. Typically two of the addresses represent source operands or arguments of the operation and the third represents the result. Some of the 3-address operations have fewer than three addresses; we simply think of the missing addresses as unused (or ignored) fields in the instruction.

Possible Address Types

1. (Source program) Names. Really the intermediate code would contain a reference to the (identifier) table entry for the name. For convenience, the actual identifier is often written. An important issue is type conversion, which will be discussed later.
2. Constants. Again, this would often be a reference to a table entry. As with names type conversion is an important issue for constants.
3. (Compiler-generated) Temporaries. Although it may at first seem wasteful, modern practice assigns a new name to each temporary, rather than reusing the same temporary when possible. (Remember that a DAG node is considered one temporary even if it has many parents.) Later phases can combine several temporaries into one (e.g., if they have disjoint lifetimes).

L-values and R-values

Consider
      Q := Z;     or     A[f(x)+B*D] := g(B+C*h(x,y));.
Where [] indicates array reference and () indicates a function call.

From a macroscopic view, performing each assignment involves three tasks.

1. Evaluate the left hand side (LHS) to obtain an l-value.
2. Evaluate the RHS to obtain an r-value.
3. Perform the assignment.

Note the differences between L-values, quantities that can appear on the LHS of an assignment, and and R-values, quantities that can appear only on the RHS.

• An l-value corresponds to an address or a location.
• An r-value corresponds to a value.
• Neither 12 nor s+t can be used as an l-value, but both are legal r-values.
• We say that the l-value can be dereferenced to obtain the corresponding r-value.

Possible three-address instructions

There is no universally agreed to set of three-address instructions, or even whether 3-address code should be the intermediate code for the compiler. Some prefer a set close to a machine architecture. Others prefer a higher-level set closer to the source, for example, subsets of C have been used. Others prefer to have multiple levels of intermediate code in the compiler and define a compilation phase that converts the high-level intermediate code into the low-level intermediate code. What follows is the set proposed in the book.

In the list below, x, y, and z are addresses; i is an integer, not an address); and L is a symbolic label. The instructions can be thought of as numbered and the labels can be converted to the numbers with another pass over the output or via backpatching, which is discussed below.

1. Binary and unary ops.
   1. x = y op z     with op a binary operation.
   2. x = op y       with op a unary operation.
   3. x = y           another unary operation, specifically the identity f(x)=x.
   
   
2. Binary, unary, and nullary op jumps
   1. Nullary Junp.
      goto L
   2. Conditional unary op jumps.
      if x goto L
      ifFalse x goto L.
   3. Conditional binary op jumps.
      if x relop y goto L
   
   
3. Procedure/Function Calls and Returns.
         param x     call p,n     y = call p,n     return     return y.
   The value n gives the number of parameters, which is needed when the argument of one function is the value returned by another, for example A = F(S,G(U,V),W) might become

   	param S
   	param U
   	param V
   	t = call G,2
   	param t
   	param W
   	A = call F,3
         

4. Indexed Copy ops. x = y[i]   x[i] = y.
   In the second example, x[i] is the address that is the-value-of-i locations after the address x; in particular x[0] is the same address as x. Similarly for y[i] in the first example. But a crucial difference is that, since y[i] is on the RHS, the address is dereferenced and the value in the address is what is used.

   Note that x[i] = y[j] is not permitted as that requires 4 addresses. x[i] = y[i] could be permitted, but is not.

5. Address and pointer ops. x = &y   x = *y   *x = y.
   x = &y sets the r-value of x equal to the l-value of y.
   x = *y sets the r-value of x equal to the contents of the location given by the r-value of y.
   *x = y sets the location given by the r-value of x equal to the r-value of y.

6.2.2: Quadruples (Quads)

Quads are an easy, almost obvious, way to represent the three address instructions: put the op into the first of four fields and the three addresses into the remaining three fields. Some instructions do not use all the fields. Many operands will be references to entries in tables (e.g., the identifier table).

6.2.3: (Indirect) Triples

Triples

A triple optimizes a quad by eliminating the result field of a quad since the result is often a temporary.

When this result occurs as a source operand of a subsequent instruction, the source operand is written as the value-number of the instruction yielding this result (distinguished some way, say with parens).

If the result field of a quad is a program name and not a temporary then two triples may be needed:

1. Do the operation and place the result into a temporary (which is not a field of this instruction).
2. A copy instruction from the temporary to the final home. Recall that a copy does not use all the fields of a quad so fits into a triple without omitting the result.

Indirect Triples

When an optimizing compiler reorders instructions for increased performance, extra work is needed with triples since the instruction numbers, which have changed, are used implicitly. Hence the triples must be regenerated with correct numbers as operands.

With Indirect triples we maintain an array of pointers to triples and, if it is necessary to reorder instructions, just reorder these pointers. This has two advantages.

1. The pointers are (probably) smaller than the triples so faster to move. This is a generic advantage and could be used for quads and many other reordering applications (e.g., sorting large records).
2. Triples contain references to results computed by prior triples. These references are given as the number of the computing triple. Since, with indirect triples, the triples themselves don't move when the instructions are reordered, the references they contain remain accurate. This advantage is specific to triples (or similar situations).

Homework: 1, 2 (you may use the parse tree instead of the syntax tree if you prefer).

6.3: Types and Declarations

Much of the early part of this section is really about programming languages more than about compilers.

6.3.1: Type Expressions

A type expression is either a basic type or the result of applying a type constructor.

6.3.2: Type Equivalence

There are two camps, name equivalence and structural equivalence.

Consider the following example.

    declare
       type MyInteger is new Integer;
       MyX : MyInteger;
       x   : Integer := 0;
    begin
       MyX := x;
    end
  

As another example, consider an object of an anonymous type as in
      X : array [5] of integer;
X does not have the same type as any other object not even Y declared as
      y : array [5] of integer;
However, x[2] has the same type as y[3]; both are integers.

6.3.3: Declarations

The following example from the 2e uses an usual C/Java-like array notation. (The 1e had pascal-like notation.) Although I prefer Ada-like constructs as in lab 3, I realize that the class knows C/Java best so like the authors I will sometimes follow the 2e as well as presenting lab3-like grammars. I will often refer to lab3-like grammars as the class grammar.

The grammar below gives C/Java-like records/structs/methodless-classes as well as multidimensional arrays (really singly dimensioned arrays of singly dimensioned arrays).

    D → T id ; D | ε
    T → B C | RECORD { D }
    B → INT | FLOAT
    C → [ NUM ] C | ε
  

Note that an example sentence derived from D is

    int [5] x ;
  

The class grammar doesn't support records. This part of the class grammar declares ints, reals, arrays, and user-defined types.

    declarations         → declaration declarations | ε
    declaration          → defining-identifier : type ; |
                           TYPE defining-identifier IS type ;
    defining-identifier  → IDENTIFIER
    type                 → INT | REAL | ARRAY [ NUMBER ] OF type | IDENTIFIER
  

For now we ignore the second possibility for a declaration (declaring a type itself).

    ds   → d ds | ε
    d    → di : ty ;
    di   → ID
    ty   → INT | REAL | ARRAY [ NUMBER ] OF ty
  

      type vector5 is array [5] of real;
      v5 : vector5;
    

      ds   → d ds | ε
      d    → di : ty ; | TYPE di IS ty ;
      di   → ID
      ty   → INT | REAL | ARRAY [ NUMBER ] OF ty | ID
    

      type t1 is array [5] of integer;
    

      type t2 is array of integer;
    

      x1 : t1
      x2 : t2[5]
    

6.3.4: Storage Layout for Local Names

Previously we considered an SDD for arrays that was able to compute the type. The key point was that it called the function array(size,type) so as to produce a tree structure exhibiting the dimensionality of the array. For example the tree on the right would be produced for
    int[3][4]     or     array [3] of array [4] of int.

Now we will extend the SDD to calculate the size of the array as well. For example, the array pictured has size 48, assuming that each int has size 4. When we declare many objects, we need to know the size of each one in order to determine the offsets of the objects from the start of the storage area.

We are considering here only those types for which the storage requirements can be computed at compile time. For others, e.g., string variables, dynamic arrays, etc, we would only be reserving space for a pointer to the structure; the structure itself would be created at run time. Such structures are discussed in the next chapter.

The idea (for arrays whose size can be determined at compile time) is that the basic type determines the width of the object, and the number of elements in the array determines the height. These are then multiplied to get the size (area) of the object. The terminology actually used is that the basetype determines the basewidth, which when multiplied by the number of elements gives the width.

The book uses semantic actions, i.e., a syntax directed translation or SDT. I added the corresponding semantic rules so that the table to the right is an SDD as well. In both cases we follow the book and show a single type specification T rather than a list of object declarations D. The omitted productions are

    D → T ID ; D | ε
  

The goal of the SDD is to calculate two attributes of the start symbol T, namely T.type and T.width, the rules can be viewed as the implementation.

The basetype and hence the basewidth are determined by B, which is INT or FLOAT. The values are set by the lexer (as attributes of FLOAT and INT) or, as shown in the table, are constants in the parser. NUM.value is set by the lexer.

These attributes of B pushed down the tree via the inherited attributes C.bw and C.bw until all the array dimensions have been passed over and the ε-production is reached. There they are turned around as usual and sent back up, during which the dimensions are processed and the final type and width are calculated.

An alternative implementation, using global variables, is described in the grayed out description of the semantic actions This is similar to the comment above that instead of having the identifier table passed up and down via attributes, the bullet can be bitten and a globally visible table used instead.

Using the Class Grammar

This exercise is easier with the class grammar since there are no inherited attributes. We again assume that the lexer has defined NUM.value (it is likely a field in the numbers table entry for the token NUM). The goal is to augment the identifier table entry for ID to include the type and size information found in the declaration. This can be written two ways.

1. addType(ID.entry,ty.type)
   addSize(ID.entry,ty.size)
2. ID.entry.type=ty.type
   ID.entry.size=ty.size

Recall that addType is viewed as synthesized since its parameters come from the RHS, i.e., from children of this node.

addType has a side effect (it modifies the identifier table) so we must ensure that we do not use this table value before it is calculated.

When do we use this value?
Answer: When we evaluate expressions, we will need to look up the types of objects.

How can we ensure that the type has already been determined and saved?
Answer: We will need to enforce declaration before use. So, in expression evaluation, we should check the entry in the identifier table to be sure that the type has already been set.

Example 1:
As a simple example, let us construct, on the board, the parse tree for the scalar declaration
      y : int ;
We get the diagram at the right, in which I have also shown the effects of the semantic rules.

Note specifically the effect of the addType and addSize functions on the identifier table.

Example 2:
For our next example we choose the array declaration
    a : array [7] of int ;
The result is again shown on the right. The green numbers show the value of t.size and the blue number shows the value of NUM.value.

Example 3:
For our final example in this section we combine the two previous declarations into the following simple complete program. In this example we show only the parse tree. In the next section we consider the semantic actions as well.

    Procedure P1 is
      y : int;
      a : array [7] of real;
    begin
    end;
  

We observe several points about this example.

1. Since we have 2 declarations, we need to use the two productions involving the nonterminal declarations (ds)
       ds → d ds | ε
2. An Euler-tree traversal of the tree will visit the declarations before visiting any statements (I know this example doesn't have any statements). Thus, if we are processing a statement and find an undeclared variable, we can signal an error since we know that there is no chance we will visit the declaration later.
3. With multiple declarations, it will be necessary to determine the offset of each declared object.
4. We will study an extension of this example in the next section. In particular, we will show the annotated parse tree, including calculations of the offset just mentioned.

6.3.5: Sequences of Declarations

Remark: Be careful to distinguish between three methods used to store and pass information.

1. Attributes. These are variables in a phase of the compiler (the semantic analyzer a.k.a intermediate code generator).
2. Identifier (and other) table. This holds longer lived data; often passed between phases.
3. Run time storage. This is storage established by the compiler, but not used by the compiler. It is allocated and used during run time.

To summarize, the identifier table (and other tables we have used) are not present when the program is run. But there must be run time storage for objects. In addition to allocating the storage (a subject discussed next chapter), we need to determine the address each object will have during execution. Specifically, we need to know its offset from the start of the area used for object storage.

For just one object, it is trivial: the offset is zero. When there are multiple objects, we need to keep a running sum of the sizes of the preceding objects, which is our next objective.

Multiple Declarations

The goal is to permit multiple declarations in the same procedure (or program or function). For C/java like languages this can occur in two ways.

1. Multiple objects in a single declaration.
2. Multiple declarations in a single procedure.

In either case we need to associate with each object being declared the location in which it will be stored at run time. Specifically we include in the table entry for the object, its offset from the beginning of the current procedure. We initialize this offset at the beginning of the procedure and increment it after each object declaration.

The lab3 grammar does not support multiple objects in a single declaration.

C/Java does permit multiple objects in a single declaration, but surprisingly the 2e grammar does not.

Naturally, the way to permit multiple declarations is to have a list of declarations in the natural right-recursive way. The 2e C/Java grammar has D which is a list of semicolon-separated T ID's
    D → T ID ; D | ε

The lab 3 grammar has a list of declarations (each of which ends in a semicolon). Shortening declarations to ds we have
    ds → d ds | ε

As mentioned, we need to maintain an offset, the next storage location to be used by an object declaration. The 2e snippet on the right introduces a nonterminal P for program that gives a convenient place to initialize offset.

The name top is used to signify that we work with the top symbol table (when we have nested scopes for record definitions, nested procedures, or nested blocks we need a stack of symbol tables). Top.put places the identifier into this table with its type and storage location and then bumps offset for the next variable or next declaration.

Rather than figure out how to put this snippet together with the previous 2e code that handled arrays, we will just present the snippets and put everything together on the class grammar.

Start Lecture #10

Multiple Declarations in the Class Grammar

On the right we show the part of the SDD used to translate multiple declarations for the class grammar. We do not show the productions for name-and-parameters (np), statements (ss), or statement (s) since we are focusing just on the declaration of local variables.

The new part is determining the offset for each individual declaration. The new items have blue backgrounds (this includes new, inherited attributes, which were red). The idea is that the first declaration has offset 0 and the next offset is the current offset plus the current size. Specifically, we proceed as follows.

In the procedure-def (pd) production, we give the nonterminal declarations (ds) the inherited attribute offset (ds.offset), which we initialize to zero.

We inherit this offset down to individual declarations. At each declaration (d), we store the offset in the entry for the identifier being declared and increment the offset by the size of this object.

When we get the to the end of the declarations (the ε-production), the offset value is the total size needed. We turn it around and send it back up the tree in case the total is needed by some higher level production.

Example: What happens when the following program (an extension of P1 from the previous section) is parsed and the semantic rules above are applied.

  procedure P2 is
      y : integer;
      a : array [7] of real;
  begin
      y := 5;      // Statements not yet done
      a[2] := y;   // Type error?
  end;

On the right is the parse tree, limited to the declarations.

The dotted lines would be solid for a parse tree and connect to the lines below. They are shown dotted so that the figure can be compared to the annotated parse tree immediately below, in which the attributes and their values have been filled in.

In the annotated tree, the attributes shown in red are inherited. Those in black are synthesized.

To start the annotation process, look at the top production in the parse tree. It has an inherited attributed ds.offset, which is set to zero. Since the attribute is inherited, the entry can be placed immediately into the annotated tree.

We now do the left child and fill in its inherited attribute.

When the Euler tour comes back up the tree, the synthesized attributes are evaluated and recorded.

6.3.6: Fields in Records and Classes

Since records can essentially have a bunch of declarations inside, we only need add

    T → RECORD { D }
  

  T → record {  { Env.push(top);
                  top = new Env();
                  Stack.push(offset);
                  offset = 0; }
  D }           { T.type = record(top);
                  T.width = offset;
                  top = Env.pop();
                  offset = Stack.pop(); }

This does not apply directly to the class grammar, which does not have records.

This same technique would be used for other examples of nested scope, e.g., nested procedures/functions and nested blocks. To have nested procedures/functions, we need other alternatives for declaration: procedure/function definitions. Similarly if we wanted to have nested blocks we would add another alternative to statement.

  s           → ks | ids | block-stmt
  block-stmt  → DECLARE ds BEGIN ss END ;

If we wanted to generate code for nested procedures or nested blocks, we would need to stack the symbol table as done above and in the text.

Homework: 1.

6.4: Translation of Expressions

The goal is to generate 3-address code for scalar expressions, i.e., arrays are not treated in this section (they will be shortly). Specifically, indices is assumed to be ε. We use is to abbreviate indices.

We generate the 3-address code using the natural notation of 6.2. In fact we assume there is a function gen() that, given the pieces needed, does the proper formatting so gen(x = y + z) will output the corresponding 3-address code. gen() is often called with addresses other than lexemes, e.g., temporaries and constants. The constructor Temp() produces a new address in whatever format gen needs. Hopefully this will be clear in the table to the right and the others that follow.

6.4.1: Operations Within Expressions

We will use two attributes code and address (addr). The key objective at each node of the parse tree for an expression is to produce values for the code and addr attributes so that the following crucial invariant is maintained.

If code is executed, then address addr contains the value of the (sub-)expression routed at this node.

Said another way, the attribute addr at a node is the address that holds the value calculated by the code at the node. Recall that unlike real code for a real machine our 3-address code doesn't reuse temporary addresses.

As one would expect for expressions, all the attributes in the table to the right are synthesized. The table is for the expression part of the class grammar. To save space we use ID for IDENTIFIER, e for expression, t for term, and f for factor.

The SDDs for declarations and scalar expressions can be easily combined by essentially concatenating them as shown here.

6.4.3: Addressing Array Elements

The idea is to associate the base address with the array name. That is, the offset stored in the identifier table entry for the array is the address of the first element. When an element is referenced, the indices and the array bounds are used to compute the amount, often called the offset (unfortunately, we have already used that term), by which the address of the referenced element differs from the base address.

To implement this technique, we must first store the base type of each identifier in the identifier table. We use this basetype to determine the size of each element of the array. For example, consider

    arr: array [10] of integer;
    x  : real ;
  

This SDD is combined with the expression SDD here.

One Dimensional Arrays

Calculating the address of an element of a one dimensional array is easy. The address increment is the width of each element times the index (assuming indices start at 0). So the address of A[i] is the base address of A, which is the offset component of A's entry in the identifier table, plus i times the width of each element of A.

The width of each element of an array is the width of what we have called the base type of the array. For a scalar, there is just one element and its width is the width of the type, which is the same as the base type. Hence, for any ID the element width is sizeof(getBaseType(ID.entry)).

For convenience, we define getBaseWidth by the formula

    getBaseWidth(ID.entry) = sizeof(getBaseType(ID.entry)) = sizeof(ID.entry.baseType)
  

Two Dimensional Arrays

Let us assume row major ordering. That is, the first element stored is A[0,0], then A[0,1], ... A[0,k-1], then A[1,0], ... . Modern languages use row major ordering.

With the alternative column major ordering, after A[0,0] comes A[1,0], A[2,0], ... .

For two dimensional arrays the address of A[i,j] is the sum of three terms

1. The base address of A.
2. The distance from A to the start of row i. This is i times the width of a row, which is i times the number of elements in a row times the width of an element. The number of elements in a row is the column array bound.
3. The distance from the start of row i to element A[i,j]. This is j times the width of an element.

Remarks

1. Our grammar really declares one dimensional arrays of one dimensional arrays rather than 2D arrays. I believe this makes it easier.
2. The SDD above when processing the declaration
       A : array [5] of array [9] of real;
   gives A the type array(5,array(9,real)) and thus the type component of the entry for A in the symbol table contains all the values needed to compute the address of any given element of the array.

End of Remarks

Higher Dimensional Arrays

The generalization to higher dimensional arrays is clear.

A Simple Example

Consider the following expression containing a simple array reference, where a and c are integers and b is a real array.

    a = b[3*c]
  

    T1 = 3 * c     // i.e. mult T1,3,c
    T2 = T1 * 8    // each b[i] is size 8
    a  = b[T2]     // Uses the x[i] special form
  

    T1 = 3 * c
    T2 = T1 * 8
    T3 = &b
    T4 = T2 + T3
    a  = *T4
  

6.4.4: Translation of Array References (Within Expressions)

To permit arrays in expressions, we need to specify the semantic actions for the production

    factor → IDENTIFIER indices
  

One-Dimensional Array References

As a warm-up, lets start with references to one-dimensional arrays. That is, instead of the above production, we consider the simpler

    factor → IDENTIFIER index
  

The table on the right does this in two ways, both with and without using the special addressing form x[j]. In the table the nonterminal index is abbreviated i . I included the version without the x[j] special form for two reasons.

1. Since we are restricted to one dimensional arrays, the full code generation for the address of an element is not hard.
2. I thought it would be instructive to see the full address generation without hiding some of it under the covers. It was definitely instructive for me!

Note that by avoiding the special form b=x[j], I ended up using two other special forms.
Is it possible to avoid the special forms?

Our Special Forms for Addressing

We just (optionally) saw an exception to the basic lisp evaluation rule. A similar exception occurs with x[j] in our three-address code. It is a special form in that, unlike the normal rules for three-address code, we don't use the address of j but instead its value. Specifically the value of j is added to the address of x.

The rules for addresses in 3-address code also include

    a = &b
    a = *b
    *a = b
  

Incorporating 1D Arrays in the Expression SDD

Our current SDD includes the production

    f → ID is
  

    f → ID
    f → ID i
  

The semantic rules for each case were given in previous tables. The ensemble to date is given here.

On the board evaluate e.code for the RHS of the simple example above: a=b[3*c].

This is an exciting moment. At long last we really seem to be compiling!

Multidimensional Array References

As mentioned above in the general case we must process the production

    f → IDENTIFIER is
  

Following the method used in the 1D case we need to construct is.code. The basic idea is shown here

The Left-Hand Side

Now that we can evaluate expressions (including one-dimensional array reverences) we need to handle the left-hand side of an assignment statement (which also can be an array reference). Specifically we need semantic actions for the following productions from the class grammar.

    id-statement   → ID rest-of-assign
    rest-of-assign → = expression ;
    rest-of-assign → indices = expression
  

Scalars and One Dimensional Arrays on the Left Hand Side

Once again we begin by restricting ourselves to one-dimensional arrays, which corresponds to replacing indices by index in the last production. The SDD for this restricted case is shown on the right.

The first three productions reduce statements (ss) and statement (s) to identifier-statement (ids), which is used for statements such as assignment and procedure invocation, that begin with an identifier. The corresponding semantic rules simply concatenate all the code produced into the top ss.code.

The identifier-statement production captures the ID and sends it down to the appropriate rest-of-assignment (ra) production where the necessary code is generated and passed back up.

The simple ra → := e; production generates ra.code by simply appending to the evaluation of the RHS, the natural assignment with ra as the LHS.

It is instructive to compare ra.code for the ra → i := e production with f.code for the f → ID i production in the expression SDD. Both compute the same offset (index*elementSize) and both use a special form for the address x[j]=y in this section and y=x[j] for expressions.

Incorporating these productions and semantics gives this SDD. Note that we have added a semantic rule to the procedure-def (pd) production. This rule simply concatenates the code in the s and ss symbols appearing between BEG and END

Multi-dimensional Arrays on the Left Hand Side

The idea is the same as when a multidimensional array appears in an expression. Specifically,

1. Traverse the sub-tree routed at the top indices node and compute the total offset from the start of the array.
2. Multiply this offset by the width of each entry.
3. A[product] = RHS

Our Simple Example Revisited

Recall the program we could partially handle.

  procedure P2 is
      y : integer;
      a : array [7] of real;
  begin
      y := 5;      // Statements not yet done
      a[2] := y;   // Type error?
  end;

Now we can do the statements.

Homework: What code is generated for the program written above? Please remind me to go over this homework next class.

What should we do about the possible type error?

1. We could ignore errors.
2. We could assume the intermediate language permits mismatched types. Final code generation would then need to generate conversion code or signal an error.
3. We could change the program to use only one type.
4. We could learn about type checking and conversions.

Let's take the last option.

6.5: Type Checking

Type Checking includes several aspects.

1. The language comes with a type system, i.e., a set of rules saying what types can appear where.
2. The compiler assigns type expressions to parts of the source program.
3. The compiler ensures that the type usage in the program conforms to the type system for the language. This does not mean that the necessary type checks are performed at compile time.

For any language, all type checking could be done at run time, i.e. there would be no compile-time checking. However, that does not mean the compiler is absolved from the type checking. Instead, the compiler generates code to do the checks at run time.

It is normally preferable to perform the checks at compile time, if possible. Some languages have very weak typing; for example, variables can change their type during execution. Often these languages need run-time checks. Examples include lisp, snobol, and apl.

A sound type system guarantees that all checks can be performed prior to execution. This does not mean that a given compiler will make all the necessary checks.

An implementation is strongly typed if compiled programs are guaranteed to run without type errors.

6.5.1: Rules for Type Checking

There are two forms of type checking.

1. We will learn type synthesis where the types of parts are used to infer the type of the whole. For example, integer+real=real.
2. Type inference is very slick. The type of a construct is determined from its usage. This permits languages like ML to check types even though names need not be declared.

We will implement type checking for expressions. Type checking statements is similar. The SDDs below (and for lab 4) contain type checking (and coercions) for assignment statements as well as expressions.

6.5.2: Type Conversions

A very strict type system would do no automatic conversion. Instead it would offer functions for the programmer to explicitly convert between selected types. With such a system, either the program has compatible types or is in error. Such explicit conversions supplied by the programmer are called casts.

We, however, will consider a more liberal approach in which the language permits certain implicit conversions that the compiler is to supply. This is called type coercion.

We continue to work primarily with the two basic types integer and real, and postulate a unary function denoted (real) that converts an integer into the real having the same value. Nonetheless, we do consider the more general case where there are multiple types some of which have coercions (often called widenings). For example in C/Java, int can be widened to long, which in turn can be widened to float as shown in the figure to the right.

Mathematically the hierarchy on the right is a partially order set (poset) in which each pair of elements has a least upper bound (LUB). For many binary operators (all the arithmetic ones we are considering, but not exponentiation) the two operands are converted to the LUB. So adding a short to a char, requires both to be converted to an int. Adding a byte to a float, requires the byte to be converted to a float (the float remains a float and is not converted).

Start Lecture #11

Remark: Lab 4 assigned.

Checking and Coercing Types for Basic Arithmetic

The steps for addition, subtraction, multiplication, and division are all essentially the same:

1. Determine the result type, which is the LUB of the two operand types.
2. Convert each operand to the result type. This step may require generating code or may be a no op.
3. Perform the arithmetic on the converted types.

Two functions are convenient.

1. LUB(t1,t2) returns the type that is the LUB of the two given types. It signals an error if there is no LUB, for example if one of the types is an array.
2. widen(a,t,w,newcode,newaddr). Given an address a of type t, and a (hopefully) wider address w, produce the instructions newcode needed so that the address newaddr is the conversion of address a to type w.

LUB is simple, just look at the address latice. If one of the type arguments is not in the lattice, signal an error; otherwise find the lowest common ancestor. For our case the lattice is trivial, real is above int.

The widen function is more interesting. It involves n² cases for n types. Many of these are error cases (e.g., if t wider than w). Below is the code for our situation with two possible types integer and real. The four cases consist of 2 nops (when t=w), one error (t=real; w=integer) and one conversion (t=integer; w=real).

    widen (a:addr, t:type, w:type, newcode:string, newaddr:addr)
      if t=w
        newcode = ""
        newaddr = a
      else if t=integer and w=real
        newaddr = new Temp()
        newcode = gen(newaddr = (real) a)
      else signal error
  

With these two functions it is not hard to modify the rules to catch type errors and perform coercions for arithmetic expressions.

1. Maintain the type of each operand by defining type attributes for e, t, and f.
2. Coerce each operand to the LUB.

This requires that we have type information for the base entities, identifiers and numbers. The lexer supplies the type of the numbers.

It is more interesting for the identifiers. We inserted that information when we processed declarations. So we now have another semantic check: Is the identifier declared before it is used?

We will not explicitly perform this check. To do so would not be hard.

The following analysis is similar to the one given in the previous homework solution. I repeat the explanation here because it is important to understand the workings of SDDs.

Before taking on the entire SDD, let's examine a particularly interesting entry

    identifier-statement → IDENTIFIER rest-of-assignment
  

    rest-of-assignment → index := expression ;
  

Consider the assignment statement

    A[3/X+4] := X*5+Y;
  

    ra → i := e ;
  

When the tree traversal gets to the ra node the first time, its parent has passed in the value of the inherited attribute ra.id=id.entry. Thus the ra node has access to the identifier table entry for ID, which in our example is the variable A.

Prior to doing its calculations, the ra node invokes its children and gets back all the synthesized attributes. Alternately said, when the tree traversal gets to this node the last time, the children have returned all the synthesized attributes. To summarize, when the ra node finally performs its calculations, it has available.

1. ra.id: the identifier entry for A.
   
   
2. i.addr/i.code: executing i.code results in i.addr containing the value of the index, which is the expression 3/X+4.
   
   
3. e.addr/e.code: executing e.code results in e.addr containing the value of the expression X*5+Y.
   
   
4. i.type/e.type: The types of the expressions.

What must the ra node do?

1. Check that i.type is int (I don't do this, but should/could).
   
   
2. Ensure execution of i.code and e.code.
   
   
3. Multiply i.addr by the base width of the array A. (We need a temporary, ra.t1, to hold the computed value).
   
   
4. Widen e to the base type of A. (We may need, and widen would then generate, a temporary ra.addr to hold the widened value).
   
   
5. Do the actual assignment of X*5+Y to A[3/X+4].

I hope the above illustration clarifies the semantic rules for the
    ra → i := e ;
production in the SDD on the right.

Because we are not considering multidimensional-arrays, the
    f → ID is
production (is abbreviates indices) is replaced by the two special cases corresponding to scalars and one-dimensional arrays, namely:
    f → ID     f → ID i

The above illustration should also help understanding the semantic rules for this last production.

The result of incorporating these rules in our growing SDD is given here.

Homework: Same question as the previous homework (What code is generated for the program written above?). But the answer is different!
Please remind me to go over the answer next class.

For lab 4, I eliminated the left recursion for expression evaluation. This was an instructive exercise for me, so let's do it here.

6.6: Control Flow

A key to the understand of control flow is the study of Boolean expressions, which themselves are used in two roles.

1. They can be computed and treated similar to integers and reals. In many programming languages once can declare Boolean variables, use the Boolean constants true and false, and construct larger Boolean expression using Boolean operators such as and and or. There are also relational operators that produce Boolean values from arithmetic operands. From this point of view, Boolean expressions are similar to the expressions we have already treated. Our previous semantic rules could be modified to generate the code needed to evaluate these expressions.
2. Boolean expressions are used in certain statements that alter the normal flow of control. It is in this regard that we have something new to learn.

6.6.1: Boolean Expressions

One question that comes up with Boolean expressions is whether both operands need be evaluated. If we are evaluating A or B and find that A is true, must we evaluate B? For example, consider evaluating

    A=0  OR  3/A < 1.2
  

This issue arises in other cases not involving Booleans at all. Consider A*F(x). If the compiler determines that A must be zero at this point, must it generate code to evaluate F(x)? Don't forget that functions can have side effects.

6.6.2: Short-Circuit (or Jumping) Code

Consider

    IF bool-expr-1 OR bool-expr-2 THEN
        then-clause
    ELSE
        else-clause
    END;
  

When we compile the if condition bool-expr-1 OR bool-expr-2 we do not use an OR operator. Instead, we generate jumps to either the true branch or the false branch. We shall see that the above source code in the comparatively simple case where bool-expr-1=x and bool-expr-2=y will generate

    if x goto L2
    goto L4
    L4:
    if y goto L2
    goto L3
    L2:
    then-clause-code
    goto L1
    L3:
    else-clause-code
    L1:  -- This label (if-stmt.next) was defined higher in the SDD tree
  

Note that the compiled code does not evaluate y if x is true. This is the sense in which it is called short-circuit code. As we have stated above, for many programming languages, it is required that we not evaluate y if x is true.

6.6.3: Flow-of-Control Statements

The class grammar has the following productions concerning flow of control statements.

    program           → procedure-def program | ε
    procedure-def     → PROCEDURE name-and-params IS decls BEGIN statement statements END ;
    statements        → statement statements | ε
    statement         → keyword-statement | identifier-statement
    keyword-statement → return-statement | while-statement | if-statement
    if-statement      → IF boolean-expression THEN statements optional-else END ;
    optional-else     → ELSE statements | ε
    while-statement   → WHILE boolean-expression DO statements END ;
  

I don't show the productions for name-and-parameters, declarations, identifier-statement, and return-statement since they do not have conditional control flow. The productions for boolean-expression will be done in the next section.

In this section we will produce an SDD for the productions above under the assumption that the SDD for Boolean expressions generates jumps to the labels be.true and be.false (depending of course on whether the Boolean expression be is true or false).

The diagrams on the right give the idea for the three basic control flow statements, if-then (upper left), if-then-else (right), and while-do (lower-left).

The table to the right gives the details via an SDD.

To enable the tables to fit I continue to abbreviate severely the names of the nonterminals. New abbreviations are ks (keyword-statement), ifs (if-statement), be (boolean-expression), oe (optional-else) and ws (while-statement). The remaining abbreviations, to be introduced next section, are bt (boolean-term) and bf (boolean-factor)

The treatment of the various *.next attributes deserves some comment. Each statement (e.g., if-statement abbreviated ifs or while-stmt abbreviated ws) is given, as an inherited attribute, a new label *.next. You can see the new label generated in the
statements → statement statements
production and then notice how it is passed down the tree from statements to statement to keyword-statement to if-statementt. The child generates a goto *.next if it wants to end (look at ifs.code in the if-statement production).

The parent, in addition to sending this attribute to the child, normally places label(*.next) after the code for the child. See ss.code in the stmts → stmt stmts production.

An alternative design would have been for the child to itself generate the label and place it as the last component of its code (or elsewhere if needed). I believe that alternative would have resulted in a clearer SDD with fewer inherited attributes. The method actually chosen does, however, have one and possibly two advantages.

1. Perhaps there is a case where it is awkward for the child to place something at the end of its code (but I don't quite see how this could be). To investigate if this second possibility occurs one might want to examine the treatment of case statements below.
   
   
2. Look at ws.code, the code attribute for the while statement. The parent (the while-stmt) does not place the ss.next label after ss.code. Instead it is placed before ss.code (actually before the while test, which is before ss.code) since right after ss.code is an unconditional goto to the while test. If we used the alternative where the child (the stmts) generated the ss.next at the end of its code, the parent would still need a goto from after ss.code to the begin label. If the child needed to jump to its end it would jump to the ss.next label and the next statement would be the jump to begin. With the scheme we are using we do not have this embarrassing jump to jump sequence.

   I must say, however, that, given all the non-optimal code we are generating, it might have been pedagogically better to use the alternative scheme and live with the jump to a jump embarrassment. I decided in the end to follow the book, but am not sure that was the right decision.

Homework: Give the SDD for a repeat statement
    REPEAT ss WHILE be END ;

6.6.4: Control-Flow Translation of Boolean Expressions

The SDD for the evaluation of Boolean expressions is given on the right. All the SDDs are assembled together here.

Recall that in evaluating the Boolean expression the objective is not simply to produce the value true or false, but instead, the goal is to generate a goto be.true or a goto be.false (depending of course on whether the Boolean expression evaluates to true or to false).

Recall as well that we must not evaluate the right-hand argument of an OR if the left-hand argument evaluates to true and we must not evaluate the right-hand argument of an AND if the left-hand argument evaluates to true.

Notice a curiosity. When evaluating arithmetic expressions, we needed to write

    expr → expr + term
  

    expr → term + expr
  

Look at the rules for the first production in the table. If bt yields true, then we are done (i.e., we want to jump to be.true). Thus we set bt.true=be.true, knowing that when bt is evaluated (look at be.code) bt will now jump to be.true.

On the other hand, if bt yields false, then we must evaluate be₁, which is why be.code has the label for bt.false right before be₁.code. In this case the value of be is the same as the value of be₁, which explains the assignments to be₁.true and be₁.false.

Do on the board the translation of

    IF   x<5  OR  x>10 AND x=y  THEN
       x := 3;
    END ;
  

We first produce the parse tree, as shown on the upper right. The tree appears quite large for such a small example. Do remember that production compilers use the (abstract) syntax tree, which is quite a bit smaller.

Notice that each e.code="" and that each e.addr is the ID or NUM directly below it. The real action begins with the three bf and one ra productions. Another point is that there are several identity productions such as be→bt.

Thus we can simplify the parse tree to the one at the lower right. This is not an official tree, I am using it here just to save some work. Basically, I eliminated nodes that would have simply copied attributes (inherited attributes are copied downward; synthesized attributes are copied upwards). I also replaced the ID, NUM, and RELOP terminals with their corresponding lexemes. Then, as mentioned above, I moved these lexemes up the tree to the expr directly above them (the only exception is the LHS of the assignment statement; the ID there is not reduced to an expr).

The Euler-tour traversal to evaluate the SDD for this parse tree begins by processing the inherited attributes at the root. The code fragment we are considering is not a full example (recall that the start symbol of the grammar is program not if-stmt). So when we start at the if-stmt node, its inherited attribute is.next will have been evaluated (it is a label, let's call it L1) and the .code computed higher in the tree will place this label right after is.code.

We go down the tree to bool-expr with be.true and .false set to new labels say L2 and L3 (these are shown in green in the diagram), and ss.next and oe.next set to ifs.next=L1 (these two .next attributes will not be used). We go down to bf (really be→bt→bf) and set .true to L2 and .false to a new label L4. We start back up with

  bf.code=""||""||gen(if x<5 goto L2)||
                  gen(goto L4)
  

Next we go up to be and down to bt setting .true and .false to L2 and L3. Then down to bf with .true=L5 .false=L3, giving

    bf.code=gen(if x>10 goto L5)||gen(goto L3)
  

    bf.code=gen(if x=y goto L2)||gen(goto L3)
  

Next we go back up to bt and synthesize

    bt.code=gen(if>10 goto L5)||gen(goto L3)||
            label L5||gen(if x=y goto L2)||gen(goto L3)
  

We complete the Boolean expression processing by going up to be and synthesizing

    be.code=gen(if x<5 goto L2)||gen(goto L4)||label(L4)||
            gen(if>10 goto L5)||gen(goto L3)||
            label L5||gen(if x=y goto L2)||gen(goto L3)
  

We have already seen assignments. You can check that s.code=gen(x:=3)

Finally we synthesize ifs.code and get (written in a more familiar style)

    if x<5 goto L2
    goto L4
    L4:
    if x>10 goto L5
    goto L3
    L5:
    if x=y goto L2
    goto L3
    L2:
    x:=3
    goto L1
    L3:
    // oe.code is empty
    L1:
  

Note that there are four extra gotos. One is a goto the next statement. Two others could be eliminated by using ifFalse. The fourth just skips over a label and empty code.

Remark: This ends the material for lab 4.

Start Lecture #12

6.6.6: Boolean Values and Jumping Code

If there are boolean variables (or variables into which a boolean value can be placed), we can have boolean assignment statements. That is we might evaluate boolean expressions outside of control flow statements.

Recall that the code we generated for boolean expressions (inside control flow statements) used inherited attributes to push down the tree the exit labels B.true and B.false. How are we to deal with Boolean assignment statements?

Two Methods for Booleans Assignment Statements: Method 1

Up to now we have used the so called jumping code method for Boolean quantities. We evaluated Boolean expressions (in the context of control flow statements) by using inherited attributes for the true and false exits (i.e., the target locations to jump to if the expression evaluates to true and false).

With this method if we have a Boolean assignment statement, we just let the true and false exits lead respectively to statements

    LHS = true
    LHS = false
  

Two Methods for Booleans Assignment Statements: Method 2

In the second method we simply treat boolean expressions as expressions. That is, we just mimic the actions we did for integer/real evaluations. Thus Boolean assignment statements like
a = b OR (c AND d AND (x < y))
just work.

For control flow statements such as

    while boolean-expression do statement-list end ;
    if boolean-expression then statement-list else statement-list end ;
  

However, as mentioned before, this is wrong.

In C and other languages if (a=0 || 1/a > f(a)) is guaranteed not to divide by zero and the above implementation fails to provide this guarantee. Thus, even if we use method 2 for implementing Boolean expressions in assignment statements, we must implement short-circuit Boolean evaluation for control flow. That is, we need to use jumping-code for control flow. The easiest solution is to use method 1, i.e. employ jumping-code for all BOOLEAN expressions.

      if
      else if
      else if
      ...
      end if
    

6.9 Intermediate Code for Procedures (and Functions)

Much of the work for procedures involves storage issues and the run time environment; this is discussed in the next chapter.

The intermediate language we use has commands for parameters, calls, and returns. Enhancing the SDD to produce this code would not be difficult, but we shall not do it.

Type Checking Procedure Calls and Function Invocations

The first requirement is to record the signature of the procedure/function definition in the symbol (or related) table. The signature of a procedure is a vector of types, one for each parameter. The signature of a function includes as well the type of the returned value.

An implementation would enhance the SDD so that the productions involving parameter(s) are treated in a manner similar to our treatment of declarations.

Procedures/Functions Defined at the Same Nesting Level

First consider procedures (or functions) P and Q both defined at the top level, which is the only case supported by the class grammar. Assume the definition of P precedes that of Q. If the body of Q (the part between BEGIN and END) contains a call to P, then the compiler can check the types because the signature of P has already been stored.

If, instead, P calls Q, then additional mechanisms are needed since the signature of Q is not available at the call site. (Requiring the called procedure to always precede the caller would preclude the case of mutual recursion, P calls Q and Q calls P.)

1. The compiler could make a preliminary pass over the parse/syntax tree during which it just populates the tables.
2. The language could be enhanced to permit both procedure/function declarations without bodies as well as the full definitions we have seen to date. Then the declarations would be placed early in the input assuring that even a one pass compiler would encounter the declaration of the called procedure/function prior to the call site (the definition might well be later in the file).

These same considerations apply if both P and Q are nested inside another procedure/function R. The difference is that the signatures of P and Q are placed in R's symbol table rather than in the top-level symbol table.

Nesting Procedures and Functions

The class grammar does not support nested scopes at all. However, the grammatical changes are small.

1. Procedure/Function definitions become another form of declaration.
2. DECLARE declarations BEGIN statements END ; becomes another form of statement.

As we have mentioned previously, procedure/function/block nesting has a significant effect on symbol table management.

We will see in the next chapter what the code generated by the compiler must do to access the non-local names that are a consequence of nested scopes.

Chapter 7: Run Time Environments

Homework: Read Chapter 7.

7.1: Storage Organization

We are discussing storage organization from the point of view of the compiler, which must allocate space for programs to be run. In particular, we are concerned with only virtual addresses and treat them uniformly.

This should be compared with an operating systems treatment, where we worry about how to effectively map these virtual addresses to real memory. For example see the discussion concerning this diagrams in my OS class notes, which illustrate an OS difficulty with the allocation method used in this course, a method that uses a very large virtual address range. Perhaps the most straightforward solution uses multilevel page tables .

Some system require various alignment constraints. For example 4-byte integers might need to begin at a byte address that is a multiple of four. Unaligned data might be illegal or might lower performance. To achieve proper alignment padding is often used.

Areas (Segments) of Memory

As mentioned above, there are various OS issues we are ignoring, for example the mapping from virtual to physical addresses, and consequences of demand paging. In this class we simply allocate memory segments in virtual memory let the operating system worry about managing real memory. In particular, we consider the following four areas of virtual memory.

1. The code (often called text in OS-speak) is fixed size and unchanging (self-modifying code is long out of fashion). If there is OS support, the text could be marked execute only (or perhaps read and execute, but not write). All other areas would be marked non-executable.
   
   
2. There is likely data of fixed size that can be determined by the compiler by examining just the program's structure without determining the program's execution pattern. One example is global data. Storage for this data would be allocated in the next area right after the code. A key point is that since the size of both the code and this so-called static area do not change during execution, these areas, unlike the next two so called dynamic areas, have no need for an expansion region.
   
   
3. The stack is used for memory whose lifetime is stack-like. It is organized into activation records that are created as a procedure is called and destroyed when the procedure exits. It abuts the area of unused memory so can grow easily. Typically the stack is stored at the highest virtual addresses and grows downward (toward small addresses). However, it is sometimes easier in describing the activation records and their uses to pretend that the addresses are increasing (so that increments are positive). We will discuss the stack in some detail below.
   
   
4. The heap is used for data whose lifetime is not as easily described. This data is allocated by the program itself, typically either with a language construct, such as new, or via a library function call, such as malloc(). It is deallocated either by another executable statement, such as a call to free(), or automatically by the system via garbage collection. We will have little more to say about the heap.

7.1.1: Static Versus Dynamic Storage Allocation

Much (often most) data cannot be statically allocated. Either its size is not known at compile time or its lifetime is only a subset of the program's execution.

Early versions of Fortran used only statically allocated data. This required that each array had a constant size specified in the program. Another consequence of supporting only static allocation was that recursion was forbidden (otherwise the compiler could not tell how many versions of a variable would be needed).

Modern languages, including newer versions of Fortran, support both static and dynamic allocation of memory.

The advantage of supporting dynamic storage allocation is the increased flexibility and storage efficiency possible (instead of declaring an array to have a size adequate for the largest data set; just allocate what is needed). The advantage of static storage allocation is that it avoids the runtime costs for allocation/deallocation and may permit faster code sequences for referencing the data.

An (unfortunately, all too common) error is a so-called memory leak where a long running program repeated allocates memory that it fails to delete, even after it can no longer be referenced. To avoid memory leaks and ease programming, several programming language systems employ automatic garbage collection. That means the runtime system itself determines when data can no longer be referenced and automatically deallocates it.

7.2: Stack Allocation of Space

The scheme to be presented achieves the following objectives.

1. Memory is shared by procedure calls that have disjoint durations. Note that we are not able to determine disjointness by just examining the program itself (due to data dependent branches among other issues).
2. The relative address of each (visible) nonlocal variable is constant throughout each execution of a procedure. Note that during this execution the procedure can call other procedures.

7.2.1: Activation Trees

Recall the fibonacci sequence 1,1,2,3,5,8, ... defined by f(1)=f(2)=1 and, for n>2, f(n)=f(n-1)+f(n-2). Consider the function calls that result from a main program calling f(5). Surrounding the more-general pseudocode that calculates (very inefficiently) the first 10 fibonacci numbers, we show the calls and returns that result from main calling f(5). On the left they are shown in a linear fashion and, on the right, we show them in tree form. The latter is sometimes called the activation tree or call tree.

  System starts main                  int a[10];
      enter f(5)                      int main(){
          enter f(4)                      int i;
              enter f(3)                  for (i=0; i<10; i++){
                  enter f(2)                a[i] = f(i);
                  exit f(2)               }
                  enter f(1)          }
                  exit f(1)           int f (int n) {
              exit f(3)                   if (n<3)  return 1;
              enter f(2)                  return f(n-1)+f(n-2);
              exit f(2)               }
          exit f(4)
          enter f(3)
              enter f(2)
              exit f(2)
              enter f(1)
              exit f(1)
          exit f(3)
      exit f(5)
  main ends

We can make the following observations about these procedure calls.

1. If an activation of p calls q, then that activation of p terminates no earlier than the activation of q.
   
   
2. The order of activations (procedure calls) corresponds to a preorder traversal of the call tree.
   
   
3. The order of de-activations (procedure returns) corresponds to postorder traversal of the call tree.
   
   
4. The Euler-tour order captures both calls and returns.
   
   
5. While executing a node N of the activation tree, the currently live activations are those corresponding to N and its ancestors in the tree.
   
   
6. These live activations were called in the order given by the root-to-N path in the tree, and the returns will occur in the reverse order.

Homework: 1, 2.

7.2.2: Activation Records (ARs)

The information needed for each invocation of a procedure is kept in a runtime data structure called an activation record (AR) or frame. The frames are kept in a stack called the control stack.

Note that this is memory used by the compiled program, not by the compiler. The compiler's job is to generate code that obtains the needed memory and to reference the data stored in the ARs.

At any point in time the number of frames on the stack is the current depth of procedure calls. For example, in the fibonacci execution shown above when f(4) is active there are three activation records on the control stack.

ARs vary with the language and compiler implementation. Typical components are described below and pictured to the right. In the diagrams the stack grows down the page.

1. The arguments (sometimes called the actual parameters). The first few arguments are often instead placed in registers.
2. The returned value. This also is often instead placed in a register (if it is a scalar).
3. The control link. These pointers connect the ARs by pointing to the AR of the caller.
4. The access link. This link is used to reference non-local variables in languages with nested procedures, an interesting challenge that we discuss in some detail below.
5. Saved status from the caller, which typically includes the return address and the machine registers. The register values are restored when control returns to the caller.
6. Data local to the procedure being activated.
7. Temporaries. For example, recall the temporaries generated during expression evaluation. Often these can be held in machine registers. When that is not possible, e.g., when there are more temporaries than registers the temporary area is used. Actually we will see in chapter 8 that only live temporaries are relevant.

The diagram on the right shows (part of) the control stack for the fibonacci example at three points during the execution. The solid lines separate ARs; the dashed lines separate components within an AR.

In the upper left we have the initial state. We show the global variable a although it is not in an activation record. It is instead statically allocated before the program begins execution (recall that the stack and heap are each dynamically allocated). Also shown is the activation record for main, which contains storage for the local variable i. Recall that local variables are near the end of the AR.

Below the initial state we see the next state when main has called f(1) and there are two activation records, one for main and one for f. The activation record for f contains space for the argument n and and also for the returned value. Recall that arguments and the return value are allocated near the beginning of the AR. There are no local variables in f.

At the far right is a later state in the execution when f(4) has been called by main and has in turn called f(2). There are three activation records, one for main and two for f. It is these multiple activations for f that permits the recursive execution. There are two locations for n and two for the returned value.

7.2.3: Calling Sequences

The calling sequence, executed when one procedure (the caller) calls another (the callee), allocates an activation record (AR) on the stack and fills in the fields. Part of this work is done by the caller; the remainder by the callee. Although the work is shared, the AR is called the callee's AR.

Since the procedure being called is defined in one place, but normally called from many places, we would expect to find more instances of the caller activation code than of the callee activation code. Thus it is wise, all else being equal, to assign as much of the work to the callee as possible.

Although details vary among implementations, the following principle is often followed: Values computed by the caller are placed before any items of size unknown by the caller. This way they can be referenced by the caller using fixed offsets. One possible arrangement is the following.

1. Place values computed by the caller at the beginning of the activation record, i.e., adjacent to AR of the caller.
   • The number of arguments may not be the same for different calls of the same function (so called varargs, e.g. printf() in C). However the (compiler of the) caller knows how many arguments there are at this call site so, where pink calls blue, the compilers knows how far the return value is from the beginning of the blue AR.
   • Since this beginning of the blue AR is the end of the pink AR (or is one location further depending on how you count), the caller knows (but only at run time) the offset of the return value location from its own stack pointer (sp, see below).
   
   
2. Fixed length items are placed next. Their sizes are known to the caller and callee at compile time. Examples of fixed length items include the links and the saved status.
   
   
3. Finally come items allocated by the callee whose size is known only at run-time, e.g., arrays whose size depends on the parameters.
   
   
4. The stack pointer sp is conveniently placed as shown in diagram. Note three consequences of this choice.
   1. The temporaries and local data are actually above the stack. This would seem even more surprising if I used the book's terminology, which is top_sp.
   2. Fixed length data can be referenced by fixed offsets (known to the intermediate code generator) from the sp.
   3. The caller knows the distance from the beginning of the callee's AR to the sp and hence can set the sp when making the call.

The picture above illustrates the situation where a pink procedure (the caller) calls a blue procedure (the callee). Also shown is Blue's AR. It is called Blue's AR because its lifetime matches that of Blue even though responsibility for this single AR is shared by both procedures.

The picture is just an approximation: For example, the returned value is actually Blue's responsibility, although the space might well be allocated by Pink. Naturally, the returned value is only relevant for functions, not procedures. Also some of the saved status, e.g., the old sp, is saved by Pink.

The picture to the right shows what happens when Blue, the callee, itself calls a green procedure and thus Blue is also a caller. You can see that Blue's responsibility includes part of its AR as well as part of Green's.

Actions During the Call

The following actions occur during a call.

1. The caller begins the process of creating the callee's AR by evaluating the arguments and placing them in the AR of the callee. (I use arguments for the caller, parameters for the callee.)
   
   
2. The caller stores the return address and the (soon-to-be-updated) sp in the callee's AR.
   
   
3. The caller increments sp so that instead of pointing into its AR, it points to the corresponding point in the callee's AR.
   
   
4. The callee saves the registers and other (system dependent) information.
   
   
5. The callee allocates and initializes its local data.
   
   
6. The callee begins execution.

Actions During the Return

When the procedure returns, the following actions are performed by the callee, essentially undoing the effects of the calling sequence.

1. The callee stores the return value (if the callee is a function). Note that this address can be determined by the caller using the old (soon-to-be-restored) sp.
   
   
2. The callee restores sp and the registers.
   
   
3. The callee jumps to the return address.

Note that varagrs are supported.

Also note that the values written during the calling sequence are not erased and the space is not explicitly reclaimed. Instead, the sp is restored and, if and when the caller makes another call, the space will be reused.

7.2.4: Variable-Length Data on the Stack

There are two flavors of variable-length data.

• Data obtained by malloc/new have hard to determine lifetimes and are stored in the heap instead of the stack.
• Data, such as arrays with bounds determined by the parameters are still stack like in their lifetimes (if A calls B, these variables of A are allocated before and released after the corresponding variables of B).

It is the second flavor that we wish to allocate on the stack. The goal is for the callee to be able to access these arrays using addresses determined at compile time even though the size of the arrays is not known until the program is called, and indeed often differs from one call to the next (even when the two calls correspond to the same source statement).

The solution is to leave room for pointers to the arrays in the AR. These pointers are fixed size and can thus be accessed using offsets known at compile time. When the procedure is invoked and the sizes are known, the pointers are filled in and the space allocated.

A difficulty caused by storing these variable size items on the stack is that it no longer is obvious where the real top of the stack is located relative to sp. Consequently another pointer (we might call it real-top-of-stack) is also kept. This is used when the callee itself calls another procedure to tell where the new allocation record should begin.

Homework: 4.

7.3: Access to Nonlocal Data on the Stack

As we shall see, the ability of procedure P to access data declared outside of P (either declared globally outside of all procedures or, especially, those declared inside another procedure Q with P nested inside Q) offers interesting challenges.

7.3.1: Data Access Without Nested Procedures

In languages like standard C without nested procedures, visible names are either local to the procedure in question or are declared globally.

1. For global names the address is known statically at compile time, providing there is only one source file. If there are multiple source files, the linker knows. In either case no reference to the activation record is needed; the addresses are known prior to execution.
2. For names local to the current procedure, the address needed is in the AR at a known-at-compile-time constant offset from the sp. In the case of variable size arrays, the constant offset refers to a pointer to the actual storage.

7.3.2: Issues With Nested Procedures

With nested procedures a complication arises. Say g is nested inside f. So g can refer to names declared in f. These names refer to objects in the AR for f. The difficulty is finding that AR when g is executing. We can't tell at compile time where the (most recent) AR for f will be relative to the current AR for g since a dynamically-determined (i.e., unknown at compile time) number of routines could have been called in the middle.

There is an example in the next section. in which g refers to x, which is declared in the immediately outer scope (main) but the AR is 2 away because f was invoked in between. (In that example you can tell at compile time what was called in what order, but with a more complicated program having data-dependent branches, it is not possible.)

7.3.3: A language with Nested Procedure Declarations

The book asserts (correctly) that C doesn't have nested procedures so introduces ML, which does (and is quite slick). However, many of you don't know ML and I haven't used it. Fortunately, a common extension to C is to permit nested procedures. In particular, gcc supports nested procedures. To check my memory I compiled and ran the following program.

  #include <stdio.h>

  int main (int argc, char *argv[]) {
      int x = 10;

      int g(int y) {
	  int z = x+y;
	  return z;
      }

      int f (int y) {
	  return g(2*y);
      }

  (void) printf("The answer is %d\n", f(x));
  return 0;
  }

The program compiles without errors and the correct answer of 30 is printed.

So we can use C (really the GCC extension of C).

Remark: Many consider this gcc extension (or its implementation) to be evil. http://www.mailinglistarchive.com/xcode-users@lists.apple.com/msg14135.html http://lkml.indiana.edu/hypermail/linux/kernel/0405.1/1008.html http://lkml.indiana.edu/hypermail/linux/kernel/0405.1/0978.html

7.3.4: Nesting Depth

Outermost procedures have nesting depth 1. Other procedures have nesting depth 1 more than the nesting depth of the immediately outer procedure. In the example above, main has nesting depth 1; both f and g have nesting depth 2.

7.3.5: Access Links

The AR for a nested procedure contains an access link that points to the AR of the most recent activation of the immediately outer procedure).

So in the example above the access link for f and the access link for g would each point to the AR of the activation of main. Then when g references x, defined in main, the activation record for main can be found by following the access link in the AR for f. Since f is nested in main, they are compiled together so, once the AR is determined, the same techniques can be used as for variables local to f.

This example was too easy.

1. Everything can be determined at compile time since there are no data dependent branches.
2. This is only one AR for main during all of execution since main is not (directly or indirectly) recursive and there is only one AR for each of f and g.

However the technique is quite general. For a procedure P to access a name defined in the 3-outer scope, i.e., the unique outer scope whose nesting depth is 3 less than that of P, you follow the access links three times. Make sure you understand why the n-th outer scope, but not the n-th inner scope is unique)

The remaining question is How are access links maintained?.

7.3.6: Manipulating Access Links

Let's assume there are no procedure parameters. We are also assuming that the entire program is compiled at once.

Without procedure parameters, the compiler knows the name of the called procedure and hence its nesting depth. The compiler always knows the nesting depth of the caller.

Let the caller be procedure F and let the called procedure be G, so we have F calling G. Let N(proc) be the nesting depth of the procedure proc.

We distinguish two cases.

1. N(G)>N(F). The only way G can be visible in F and have a greater nesting depth is for G to be declared immediately inside F. Then when compiling the call from F to G it is easy to set the access link of G to point to the AR of F. In this case the access link is the same as the control link.
   
   
2. N(G7)≤N(F). This includes the case F=G, i.e., a direct recursive call. For G to be visible in F, there must be another procedure P enclosing both F and G, with G immediately inside P. That is, we have the following situation, where k=N(G)-N(F)≥0

   	P() {
             G() {...}
             P1() {
               P2() {
                 ...
                 Pk() {
                   F(){... G(); ...}
                 }
                ...
               }
             }
           }
         

   Our goal while creating the AR for G at the call from F is to set the access link to point to the AR for P. Note that the entire structure in the skeleton code shown is visible to the compiler. The current (at the time of the call) AR is the one for F and, if we follow the access links k times we get a pointer to the AR for P, which we can then place in the access link for the being-created AR for G.

The above works fine when F is nested (possibly deeply) inside G. It is the picture above but P1 is G.

When k=0 we get the gcc code I showed before and also the case of direct recursion where G=F. I do not know why the book separates out the case k=0, especially since the previous edition didn't.

7.4.5: Manual Deallocation Requests

Stack data is automatically deallocated when the defining procedure returns. What should we do with heap data explicated allocated with new/malloc?

The manual method is to require that the programmer explicitly deallocate these data. Two problems arise.

1. Memory leaks. The programmer forgets to deallocate.

           loop
               allocate X
               use X
               forget to deallocate X
   	end loop
         

   As this program continues to run it will require more and more storage even though is actual usage is not increasing significantly.
   
   
2. Dangling References. The programmer forgets that they did a deallocate.

           allocate X
           use X
           deallocate X
           100,000 lines of code not using X
           use X
         

Both can be disastrous and motivate the next topic, which is covered in programming languages courses.

7.5: Introduction to Garbage Collection

The system detects data that cannot be accessed (no direct or indirect references exist) and deallocates the data automatically.

Covered in programming languages.

Start Lecture #13

Chapter 8: Code Generation

Homework: Read Chapter 8.

Goal: Transform the intermediate code and tables produced by the front end into final machine (or assembly) code. Code generation plus optimization constitutes the back end of the compiler.

8.1: Issues in the Design of a Code Generator

8.1.1: Input to the Code Generator

As expected the input to the code generator is the output of the intermediate code generator. We assume that all syntactic and semantic error checks have been done by the front end. Also, all needed type conversions are already done and any type errors have been detected.

We are using three address instructions for our intermediate language. These instructions have several representations, quads, triples, indirect triples, etc. In this chapter I will tend to use the term quad (for brevity) when I should really say three-address instruction, since the representation doesn't matter.

8.1.2: the Target Program

A RISC (Reduced Instruction Set Computer), e.g. PowerPC, Sparc, MIPS (popular for embedded systems), is characterized by

• Many registers.
• Three address instructions.
• Simple addressing modes.
• Relatively simple ISA (instruction set architecture).
• Only loads and stores touch memory.
• Homogeneous registers.
• Very few instruction lengths.

A CISC (Complex Instruct Set Computer), e.g. x86, x86-64/amd64 is characterized by

• Few registers.
• Two address instructions.
• Variety of addressing modes (some complex).
• Complex ISA.
• Register classes.
• Multiple instruction lengths.

A stack-based computer is characterized by

• No registers.
• Zero address instructions (operands and results are implicitly on the runtime stack).
• The top portion of the stack are kept in hidden registers.

An accumulator-based computer is characterized by

• One special register (the accumulator) where results are placed.
• Other (index) registers often used in loops for the index and to help specify the address of an array element.
• One address instructions (the other operand is the accumulator).

A Little History

IBM 701/704/709/7090/7094 (Moon shot, MIT CTSS) were accumulator based.

Stack based machines were believed to be good compiler targets. They became very unpopular when it was believed that register architecture would perform better. Better compilation (code generation) techniques appeared that could take advantage of the multiple registers.

Pascal P-code and Java byte-code are the machine instructions for a hypothetical stack-based machines, the JVM (Java Virtual Machine) in the case of Java. This code can be interpreted or compiled to native code.

RISC became all the rage in the 1980s.

CISC made a gigantic comeback in the 90s with the intel pentium pro. A key idea of the pentium pro is that the hardware would dynamically translate a complex x86 instruction into a series of simpler RISC-like instructions called ROPs (RISC ops). The actual execution engine dealt with ROPs. The jargon would be that, while the architecture (the ISA) remained the x86, the micro-architecture was quite different and more like the micro-architecture seen in previous RISC processors.

Assemblers and Linkers

For maximum compilation speed of modest size programs, the compiler accepts the entire program at once and produces code that can be loaded and executed (the compilation system can include a simple loader and can start the compiled program). This was popular for student jobs when computer time was expensive. The alternative, where each procedure can be compiled separately, requires a linkage editor.

It eases the compiler's task to produce assembly code instead of machine code and we will do so. This decision increases the total compilation time since it requires an extra assembler pass (or two).

8.1.3: Instruction Selection

A big question is the level of code quality we seek to attain. For example we can simply translate one quadruple at a time. The quad
        x = y + z
can always (assuming the addresses x, y, and z are each a compile time constant off a given register, e.g., the sp) be compiled into 4 RISC-like instructions (fewer CISC instructions would suffice) using only 2 registers R0 and R1.

    LD  R0, y
    LD  R1, z
    ADD R0, R0, R1
    ST  x, R0
  

    a = b + c
    d = a + e
  

    LD  R0, b
    LD  R1, c
    ADD R0, R0, R1
    ST  a, R0
    LD  R0, a
    LD  R1, e
    ADD R0, R0, R1
    ST  d, R0
  

8.1.4: Register Allocation

Since registers are the fastest memory in the computer, the ideal solution is to store all values in registers. However, there are normally not nearly enough registers for this to be possible. So we must choose which values are in the registers at any given time.

Actually this problem has two parts.

1. Which values should be stored in registers?
2. Which register should each selected value be stored in

The reason for the second problem is that often there are register requirements, e.g., floating-point values in floating-point registers and certain requirements for even-odd register pairs (e.g., 0&1 but not 1&2) for multiplication/division. We shall concentrate on the first problem.

8.1.5: Evaluation Order

Sometimes better code results if the quads are reordered. One example occurs with modern processors that can execute multiple instructions concurrently, providing certain restrictions are met (the obvious one is that the input operands must already be evaluated).

8.2: The Target Language

This is a delicate compromise between RISC and CISC. The goal is to be simple but to permit the study of nontrivial addressing modes and the corresponding optimizations. A charging scheme is instituted to reflect that complex addressing modes are not free.

8.2.1: A Simple Target Machine Model

We postulate the following (RISC-like) instruction set

1. Load.     LD dest, addr
   loads the destination dest with the contents of the address addr.
   LD reg1, reg2
   is a register copy.

   A question is whether dest can be a memory location or whether it must be a register. This is part of the RISC/CISC debate. In CISC parlance, no distinction is made between load and store, both are examples of the general move instruction that can have an arbitrary source and an arbitrary destination.

   We will normally not use a memory location for the destination of a load (or the source of a store). This implies that we are not able to perform a memory to memory copy in one instruction.

   As will be seen below, in those places where a memory location is permitted, we charge more than for a register.

2. Store.     ST addr, src
   stores the value of the source src (register) into the address addr.

   We do permit an integer constant preceded by a number sign, (e.g., #181) to be used instead of a register, but again we charge extra.

3. Computation.     OP dest, src1, src2     or     dest = src1 OP src2
   performs the operation OP on the two source operands src1 and src2.

   For a RISC architecture the three operands must be registers. This will be our emphasis (extra charge for an integer src). If the destination register is one of the sources, the source is read first and then overwritten (in one cycle by utilizing a master-slave flip-flop, when both are registers.)

4. Unconditional branch.     BR L
   transfers control to the (instruction with) label L.

   When used with an address rather than a label it means to goto that address. Note that we are using the l-value of the address.

   Remark: This is unlike the situation with a load instruction in which case the r-value i.e., the contents of the address, is loaded into the register. Please do not be confused by this usage. The address or memory address always refers to the location, i.e., the l-value. Some instructions, e.g., LD, require that the location be dereferenced, i.e, that the r-value be obtained

5. Conditional Branch.     Bcond r, L
   transfers to the label (or location) L if register r satisfies the condition cond. For example,
         BNEG R0, joe
   branches to joe if R0 is negative.

Addressing modes

The addressing modes are not simply RISC-like, as they permit indirection through memory locations. Again, note that we shall charge extra for some such operands.

Recall the difference between an l-value and an r-value, e.g. the difference between the uses of x in
      x = y + 3
and
      z = x + 12
The first refers to an address, the second to a value (stored in that address).

We assume the machine supports the following addressing modes.

1. Variable name. This is shorthand (or assembler-speak) for the memory location containing the variable, i.e., we use the l-value of the variable name. So
       LD R1, a
   sets the contents of R1 equal to the contents of a, i.e.,
       contents(R1) := contents(a)

   Do not get confused here. The l-value of a is used as the address (that is what the addressing mode tells us). But the load instruction itself loads the first operand with the contents of the second. That is why it is the r-value of the second operand that is placed into the first operand.

2. Indexed address. The address a(reg), where a is a variable name and reg is a register (i.e., a register number), specifies the address that is the r-value-of-reg bytes past the address specified by a. That is, the address is computed as the l-value of a plus the r-value of reg. So
       LD r1, a(r2)
   sets
       contents(r1) := contents(a+contents(r2))
   NOT
       contents(r1) := contents(contents(a)+contents(r2))

   Permitting this addressing mode outside a load or store instruction, which we shall not do, would strongly suggest a CISC architecture.

3. Indexed constant. An integer constant can be indexed by a register. So
       LD r1, 8(r4)
   sets
       contents(r1) := contents(8+contents(r4)).
   In particular,
       LD r1, 0(r4)
   sets
       contents(r1) := contents(contents(r4)).
   

4. Indirect addressing. If I is an integer constant and r is a register, the previous addressing mode tells us that I(r) refers to the address I+contents(r).
   The new addressing mode *I(r) refers to the address contents(I+contents(r)).
   The address *r is shorthand for *0(r).
   The address *10 is shorthand for *10(fakeRegisterContainingZero).
   So
       LD r1, *50(r2)
   sets
       contents(r1) := contents(contents(50+contents(r2))).
   and
         LD r1, *r2
   sets (get ready)
       contents(r1) := contents(contents(contents(r2)))
   and
       LD r1, *10
   sets
       contents(r1) := contents(contents(10))

5. Immediate constant. If a constant is preceded by a # it is treated as an r-value instead of as a register number. So

   	  ADD r2, r2, #1
   	

   is an increment instruction. Indeed

   	  ADD 2, 2, #1
   	

   does the same thing, but we probably won't write that; for clarity we will normally write registers beginning with an r

Addressing Mode Usage

Remember that in 3-address instructions, the variables written are addresses, i.e., they represent l-values.

Let us assume the l-value of a is 500 and the l-value b is 700, i.e., a and b refer to locations 500 and 700 respectively. Assume further that location 100 contains 666, location 500 contains 100, location 700 contains 900, and location 900 contains 123. This initial state is shown in the upper left picture.

In the four other pictures the contents of the pink location has been changed to the contents of the light green location. These correspond to the three-address assignment statements shown below each picture. The machine instructions indicated below implement each of these assignment statements.

    a = b
    LD  R1, b
    ST  a, R1

    a = *b
    LD  R1, b
    LD  R1, 0(R1)
    ST  a, R1

    *a = b
    LD  R1, b
    LD  R2, a
    ST  0(R2), R1

    *a = *b
    LD  R1, b
    LD  R1, 0(R1)
    LD  R2, a
    ST  0(R2), R1
  

Naive Translation of Quads to Instructions

For many quads the naive (RISC-like) translation is 4 instructions.

1. Load the first source into a register.
2. Load the second source into another register.
3. Do the operation.
4. Store the result.

Array assignment statements are also four instructions. We can't have a quad A[i]=B[j] because that needs four addresses and quads have only three. Similarly, we can't use an array in a computation statement like a[i]=x+y because it again would need four addresses.

The instruction x=A[i] becomes (assuming each element of A is 4 bytes. Actually, our intermediate code generator already does the multiplication so we would not generate a multiply here).

    LD  R0, i
    MUL R0, R0, #4
    LD  R0, A(R0)
    ST  x, R0
  

Similarly A[i]=x becomes (again our intermediate code generator already does the multiplication).

    LD  R0, i
    MUL R0, R0, #4
    LD  R1, x
    ST  A(R0), R1
  

The (C-like) pointer reference x = *p becomes

    LD  R0, p
    LD  R0, 0(R0)
    ST  x, R0
  

The assignment through a pointer *p = x becomes

    LD  R0, x
    LD  R1, p
    ST  0(R1), R0
  

Finally, if x < y goto L becomes

    LD   R0, x
    LD   R1, y
    SUB  R0, R0, R1
    BNEG R0, L
  

Conclusion

With a modest amount of additional effort much of the output of lab 4 could be turned into naive assembly language. We will not do this. Instead, we will spend the little time remaining learning how to generate less-naive assembly language.

8.2.2: Program and Instruction Costs

Generating good code requires that we have a metric, i.e., a way of quantifying the cost of executing the code.

The run-time cost of a program depends on (among other factors)

• The cost of the generated instructions.
• The number of times the instructions are executed.

Here we just determine the first cost, and use quite a simple metric. We charge for each instruction one plus the cost of each addressing mode used.

Addressing modes using just registers have zero cost, while those involving memory addresses or constants are charged one. None of our addressing modes have both a memory address and a constant or two of either one.

The charge corresponds to the size of the instruction since a memory address or a constant is assumed to be stored in a word right after the instruction word itself.

You might think that we are measuring the memory (or space) cost of the program not the time cost, but this is mistaken: The primary space cost is the size of the data, not the size of the instructions. One might say we are charging for the pressure on the I-cache.

For example, LD R0, *50(R2) costs 2, the additional cost is for the constant 50.

I believe that the book special cases the addresses 0(reg) and *0(reg) so that the 0 is not explicitly stored and not charged for. The significance for us is calculating the length an instruction such as

    LD  R1, 0(R2)
  

Homework: 1, 2, 3, 4. Calculate the cost for 2c.

8.3: Address in the Target Code

There are 4 possibilities for addresses that must be generated depending on which of the following areas the address refers to.

1. The text or code area. The location of items in this area is statically determined, i.e., is known at compile time.
2. The static area holding global constants. The location of items in this area is statically determined.
3. The stack holding activation records. The location of items in this area is not known at compile time.
4. The heap. The location of items in this area is not known at compile time.

8.3.1: Static Allocation

Returning to the glory days of Fortran, we first consider a system with only static allocation, i.e., with all address in the first two classes above. Remember, that with static allocation we know before execution where all the data will be stored. There are no recursive procedures; indeed, there is no run-time stack of activation records. Instead the ARs (one per procedure) are statically allocated by the compiler.

Caller Calling Callee

In this simplified situation, calling a parameterless procedure just uses static addresses and can be implemented by two instructions. Specifically,
      call callee
can be implemented by

    ST  callee.staticArea, #here+20
    BR  callee.codeArea
  

Assume, for convenience, that the return address is the first location in the activation record (in general, for a parameterless procedure, the return address would be a fixed offset from the beginning of the AR). We use the attribute staticArea for the address of the AR for the given procedure (remember again that there is no stack and no heap).

What is the mysterious #here+20?

We know that # signifies an immediate constant. We use here to represent the address of the current instruction (the compiler knows this value since we are assuming that the entire program, i.e., all procedures, are compiled at once). The two instructions listed contain 3 constants, which means that the entire sequence takes 2+3=5 words or 20 bytes. Thus here+20 is the address of the instruction after the BR, which is indeed the return address.

Callee Returning

With static allocation, the compiler knows the address of the the AR for the callee and we are assuming that the return address is the first entry. Then a procedure return is simply

    BR  *callee.staticArea
  

The value callee.staticArea is the address of a memory location into which the caller placed the return address. So the branch is not to callee.staticArea, but instead to the return address, which is the value contained in callee.staticArea.

Note: You might well wonder why the load

    LD r0, callee.staticArea
  

The answer, as mentioned above, is that branch and load have different semantics: Both take an address as an argument, branch jumps to that address, whereas, load retrieves the contents.

Example

We consider a main program calling a procedure P and then halting. Other actions by Main and P are indicated by subscripted uses of other.

  // Quadruples of Main
  other1
  call P
  other2
  halt
  // Quadruples of P
  other3
  return

Let us arbitrarily assume that the code for Main starts in location 1000 and the code for P starts in location 2000 (there might be other procedures in between). Also assume that each other_i requires 100 bytes (all addresses are in bytes). Finally, we assume that the ARs for Main and P begin at 3000 and 4000 respectively. Then the following machine code results.

    // Code for Main
    1000: Other1
    1100: ST 4000, #1120    // P.staticArea, #here+20
    1112: BR 2000           // Two constants in previous instruction take 8 bytes
    1120: other2
    1220: HALT
    ...
    // Code for P
    2000: other3
    2100: BR *4000
    ...
    // AR for Main
    3000:                   // Return address stored here (not used)
    3004:                   // Local data for Main starts here
    ...
    // AR for P
    4000:                   // Return address stored here
    4004:                   // Local data for P starts here
  

Start Lecture #14

8.3.2: Stack Allocation

We now need to access the ARs from the stack. The key distinction is that the location of the current AR is not known at compile time. Instead a pointer to the stack must be maintained dynamically.

We dedicate a register, call it SP, for this purpose. In this chapter we let SP point to the bottom of the current AR, that is the entire AR is above the SP. Since we are not supporting varargs, there is no advantage to having SP point to the middle of the AR as in the previous chapter.

The main procedure (or the run-time library code called before any user-written procedure) must initialize SP with
      LD SP, #stackStart
where stackStart is a known-at-compile-time constant.

The caller increments SP (which now points to the beginning of its AR) to point to the beginning of the callee's AR. This requires an increment by the size of the caller's AR, which of course the caller knows.

Is this size a compile-time constant?

The book treats it as a constant. The only part that is not known at compile time is the size of the dynamic arrays. Strictly speaking this is not part of the AR, but it must be skipped over since the callee's AR starts after the caller's dynamic arrays.

Perhaps for simplicity we are assuming that there are no dynamic arrays being stored on the stack. If there are arrays, their size must be included in some way.

Caller Calling Callee

The code generated for a parameterless call is

    ADD SP, SP, #caller.ARSize
    ST  0(SP), #here+16        // save return address (book wrong)
    BR  callee.codeArea
  

Callee Returning

The return requires code from both the Caller and Callee. The callee transfers control back to the caller with
      BR *0(SP)
Upon return the caller restore the stack pointer with
      SUB SP, SP, #caller.ARSize

Example

We again consider a main program calling a procedure P and then halting. Other actions by Main and P are indicated by subscripted uses of `other'.

    // Quadruples of Main
    other1
    call P
    other2
    halt
    // Quadruples of P
    other3
    return
  

Recall our assumptions that the code for Main starts in location 1000, the code for P starts in location 2000, and each other_i requires 100 bytes. Let us assume the stack begins at 9000 (and grows to larger addresses) and that the AR for Main is of size 400 (we don't need P.ARSize since P doesn't call any procedures). Then the following machine code results.

    // Code for Main
    1000: LD  SP, 9000        // Possibly done prior to Main
    1008: Other1
    1108: ADD SP, SP, #400
    1116: ST  0(SP), #1132    // Understand the address
    1124: BR, 2000
    1132: SUB SP, SP, #400
    1140: other2
    1240: HALT
    ...
    // Code for P
    2000: other3
    2100: BR *0(SP)           // Understand the *
    ...
    // AR for Main
    9000:                     // Return address stored here (not used)
    9004:                     // Local data for Main starts here
    9396:                     // Last word of the AR is bytes 9396-9399
    ...
    // AR for P
    9400:                     // Return address stored here
    9404:                     // Local data for P starts here
    9496:                     // Last word of the AR is bytes 9496-9799
  

Homework: 1, 2, 3.

8.4: Basic Blocks and Flow Graphs

As we have seen, for many quads it is quite easy to generate a series of machine instructions to achieve the same effect. As we have also seen, the resulting code can be quite inefficient. For one thing the last instruction generated for a quad is often a store of a value that is then loaded right back in the next quad (or one or two quads later).

Another problem is that we don't make much use of the registers. That is, translating a single quad needs just one or two registers so we might as well throw out all the other registers on the machine.

Both of the problems are due to the same cause: Our horizon is too limited. We must consider more than one quad at a time. But wild flow of control can make it unclear which quads are dynamically near each other. So we want to consider, at one time, a group of quads for which the dynamic order of execution is tightly controlled. We then also need to understand how execution proceeds from one group to another. Specifically the groups are called basic blocks and the execution order among them is captured by the flow graph.

Definition: A basic block is a maximal collection of consecutive quads such that

1. Control enters the block only at the first instruction.
2. Branches (or halts) occur only at the last instruction.

Definition: A flow graph has the basic blocks as vertices and has edges from one block to each possible dynamic successor.

We process all the quads in a basic block together making use of the fact that the block is not entered or left in the middle.

8.4.1: Basic Blocks

Constructing the basic blocks is easy. Once you find the start of a block, you keep going until you hit a label or jump. But, as usual, to say it correctly takes more words.

Definition: A basic block leader (i.e., first instruction) is any of the following (except for the instruction just past the entire program).

1. The first instruction of the program.
2. A target of a (conditional or unconditional) jump.
3. The instruction immediately following a jump.

Given the leaders, a basic block starts with a leader and proceeds up to but not including the next leader.

Example

The following code produces a 10x10 real identity matrix

    for i from 1 to 10 do
      for j from 1 to 10 do
        a[i,j] = 0
      end
    end
    for i from 1 to 10 do
      a[i,i] = 1.0
    end
  

The following quads do the same thing. Don't worry too much about how the quads were generated.

     1)  i = 1
     2)  j = 1
     3)  t1 = 10 * i
     4)  t2 = t1 + j        // element [i,j]
     5)  t3 = 8 * t2        // offset for a[i,j] (8 byte reals)
     6)  t4 = t3 - 88       // program array starts at [1,1] assembler at [0,0]
     7)  a[t4] = 0.0
     8)  j = j + 1
     9)  if j <= 10 goto (3)
    10)  i = i + 1
    11)  if i <= 10 goto (2)
    12)  i = 1
    13)  t5 = i - 1
    14)  t6 = 88 * t5
    15)  a[t6] = 1.0
    16)  i = i + 1
    17)  if i <= 10 goto (13)
  

Which quads are leaders?

1 is a leader by definition. The jumps are 9, 11, and 17. So 10 and 12 are leaders as are the targets 3, 2, and 13.

The leaders are then 1, 2, 3, 10, 12, and 13.

The basic blocks are therefore {1}, {2}, {3,4,5,6,7,8,9}, {10,11}, {12}, and {13,14,15,16,17}.

Here is the code written again with the basic blocks indicated.

     1)  i = 1
    

     2)  j = 1
    
     3)  t1 = 10 * i
     4)  t2 = t1 + j            // element [i,j]
     5)  t3 = 8 * t2            // offset for a[i,j] (8 byte numbers)
     6)  t4 = t3 - 88           // we start at [1,1] not [0,0]
     7)  a[t4] = 0.0
     8)  j = j + 1
     9)  if J <= 10 goto (3)
    

    10)  i = i + 1
    11)  if i <= 10 goto (2)
    
    12)  i = 1
    

    13)  t5 = i - 1
    14)  t6 = 88 * t5
    15)  a[t6] = 1.0
    16)  i = i + 1
    17)  if i <= 10 goto (13)
    
  

We can see that once you execute the leader you are assured of executing the rest of the block in order.

8.4.2: Next Use Information

We want to record the flow of information from instructions that compute a value to those that use the value. One advantage we will achieve is that if we find a value has no subsequent uses, then it is dead and the register holding that value can be used for another value.

Assume that a quad p assigns a value to x (some would call this a def of x).

Definition: A quad q uses the value computed at p (uses the def) and x is live at q if q has x as an operand and there is a possible execution path from p to q that does not pass any other def of x.

Since the flow of control is trivial inside a basic block, we are able to compute the live/dead status and next use information at the block leader by a simple backwards scan of the quads (algorithm below).

Note that if x is dead (i.e., defined before used) on entrance to B the register containing x can be reused in B.

Computing Live/Dead and Next Use Information

Our goal is to determine whether a block uses a value and if so in which statement is it first used. The following algorithm for computing uses is quite simple.

    Initialize all variables in B as being live
    Examine the quads q of the block in reverse order.
        Assume the quad q computes x and reads y and z
        Mark x as dead; mark y and z as live and used at q
  

When the loop finishes those values that are read before being written are marked as live and their first use is noted. The locations x that are written before being read are marked dead meaning that the value of x on entrance is not used.

Those values that are neither read nor written remain simply live. They are not dead since the dynamically next basic block might use them.

Note that we have determined whether values are live/dead on entrance to the basic block. We would like to know as well if they are live/dead on exit, but that requires global flow analysis, which we do not know.

8.4.3: Flow Graphs

The nodes of the flow graph are the basic blocks, and there is an edge from P (predecessor) to S (successor) if S might follow P. More formally, such an edge is added if the last statement of P

1. is a jump to S (it must be to the leader of S) or
2. is not an unconditional jump and S immediately follows P.

Two nodes are added: entry and exit. An edge is added from entry to the first basic block, i.e. the block that has the first statement of the program as leader.

Edges to the exit are added from any block that could be the last block executed. Specifically, edges are added to exit from

1. the last block if it doesn't end in an unconditional jump.
2. any block that ends in a jump to outside the program.

The flow graph for our example is shown on the right.

8.4.4: Representing Flow Graphs

Note that jump targets are no longer quads but blocks. The reason is that various optimizations within blocks will change the instructions and we would have to change the jump to reflect this.

8.4.5: Loops

For most programs the bulk of the execution time is within loops so we want to identify these.

The flow graph on the right has three loops.

1. {B₃}, i.e., B₃ by itself.
2. {B₆}.
3. {B₂, B₃, B₄}

Homework: 1.

Remark: Nothing beyond here will be on the final.

A Word or Two About Global Flow Analysis

We are not covering global flow analysis; it is a key component of optimization and would be a natural topic in a follow-on course. Nonetheless there is something we can say just by examining the flow graphs we have constructed. For this discussion I am ignoring tricky and important issues concerning arrays and pointer references (specifically, disambiguation). You may wish to assume that the program contains no arrays or pointers.

We have seen that a simple backwards scan of the statements in a basic block enables us to determine the variables that are live-on-entry and those that are dead-on-entry. Those variables that do not occur in the block are in neither category; perhaps we should call them ignored by the block.

We shall see below that it would be lovely to know which variables are live/dead-on-exit. This means which variables hold values at the end of the block that will / will not be used. To determine the status of v on exit of a block B, we need to trace all possible execution paths beginning at the end of B. If all these paths reach a block where v is dead-on-entry before they reach a block where v is live-on-entry, then v is dead on exit for block B.

8.5: Optimization of Basic Blocks

8.5.1: The DAG Representation of Basic Blocks

The goal is to obtain a visual picture of how information flows through the block. The leaves will show the values entering the block and as we proceed up the DAG we encounter uses of these values, defs (and redefs) of values, and uses of the new values.

Formally, this is defined as follows.

1. Create a leaf for the initial value of each variable appearing in the block. (We do not know what that the value is, not even if the variable has ever been given a value).
2. Create a node N for each statement s in the block.
   1. Label N with the operator of s. This label is drawn inside the node.
   2. Attach to N those variables for which N is the last def in the block. These additional labels are drawn along side of N.
   3. Draw edges from N to each statement that is the last def of an operand used by N.
3. Designate as output nodes those N whose values are live on exit, an officially-mysterious term meaning values possibly used in another block. (Determining the live on exit values requires global, i.e., inter-block, flow analysis.)

As we shall see in the next few sections various basic-block optimizations are facilitated by using the DAG.

8.5.2: Finding Local Common Subexpressions

As we create nodes for each statement, proceeding in the static order of the statements, we might notice that a new node is just like one already in the DAG in which case we don't need a new node and can use the old node to compute the new value in addition to the one it already was computing.

Specifically, we do not construct a new node if an existing node has the same children in the same order and is labeled with the same operation.

Consider computing the DAG for the following block of code.

    a = b + c
    c = a + x
    d = b + c
    b = a + x
  

The DAG construction proceeds as follows (the movie on the right accompanies the explanation).

1. First we construct leaves with the initial values.
2. Next we process a = b + c. This produces a node labeled + with a attached and having b₀ and c₀ as children.
3. Next we process c = a + x.
4. Next we process d = b + c. Although we have already computed b + c in the first statement, the c's are not the same, so we produce a new node.
5. Then we process b = a + x. Since we have already computed a + x in statement 2, we do not produce a new node, but instead attach b to the old node.
6. Finally, we tidy up and erase the unused initial values.

You might think that with only three computation nodes in the DAG, the block could be reduced to three statements (dropping the computation of b). However, this is wrong. Only if b is dead on exit can we omit the computation of b. We can, however, replace the last statement with the simpler
      b = c.

Sometimes a combination of techniques finds improvements that no single technique would find. For example if a-b is computed, then both a and b are incremented by one, and then a-b is computed again, it will not be recognized as a common subexpression even though the value has not changed. However, when combined with various algebraic transformations, the common value can be recognized.

8.5.3: Dead Code Elimination

Assume we are told (by global flow analysis) that certain values are dead on exit. We examine each root (node with no ancestor) and delete any for which all attached variables are dead on exit. This process is repeated since new roots may have appeared.

For example, if we are told, for the picture on the right, that c and d are dead on exit, then the root d can be removed since d is dead. Then the rightmost node becomes a root, which also can be removed (since c is dead).

8.5.4: The Use of Algebraic Identities

Some of these are quite clear. We can of course replace x+0 or 0+x by simply x. Similar considerations apply to 1*x, x*1, x-0, and x/1.

Another class of simplifications is strength reduction, where we replace one operation by a cheaper one. A simple example is replacing 2*x by x+x on architectures where addition is cheaper than multiplication.

A more sophisticated strength reduction is applied by compilers that recognize induction variables (loop indices). Inside a
    for i from 1 to N
loop, the expression 4*i can be strength reduced to j=j+4 and 2^i can be strength reduced to j=2*j (with suitable initializations of j just before the loop).

Other uses of algebraic identities are possible; many require a careful reading of the language reference manual to ensure their legality. For example, even though it might be advantageous to convert
    ((a + b) * f(x)) * a
to
    ((a + b) * a) * f(x)
it is illegal in Fortran since the programmer's use of parentheses to specify the order of operations can not be violated.

Does

    a = b + c
    x = y + c + b + r
  

8.5.5: Representation of Array References

Arrays are tricky. Question: Does

    x = a[i]
    a[j] = y
    z = a[i]
  

A statement of the form x = a[i] generates a node labeled with the operator =[] and the variable x, and having children a₀, the initial value of a, and the value of i.

A statement of the form a[j] = y generates a node labeled with operator []= and three children a₀. j, and y, but with no variable as label. The new feature is that this node kills all existing nodes depending on a₀. A killed node can not received any future labels so cannot becomew a common subexpression.

Returning to our example

    x = a[i]
    a[j] = y
    z = a[i]
  

Sometimes it is not children but grandchildren (or other descendant) that are arrays. For example we might have

    b = a + 8    // b[i] is 8 bytes past a[i]
    x = b[i]
    b[j] = y
  

8.5.6: Pointer Assignment and Procedure Calls

Pointers are even trickier than arrays. Together they have spawned a mini-industry in disambiguation, i.e., when can we tell whether two array or pointer references refer to the same or different locations. A trivial case of disambiguation occurs with.

    p = &x
    *p = y
  

With no disambiguation information, we must assume that a pointer can refer to any location. Consider

    x = *p
    *q = y
  

We must treat the first statement as a use of every variable; pictorially the =* operator takes all current nodes with identifiers as arguments. This impacts dead code elimination.

We must treat the second statement as writing every variable. That is all existing nodes are killed, which impacts common subexpression elimination.

In our basic-block level approach, a procedure call has properties similar to a pointer reference: For all x in the scope of P, we must treat a call of P as using all nodes with x attached and also killing those same nodes.

8.5.7: Reassembling Basic Blocks From DAGs

Now that we have improved the DAG for a basic block, we need to regenerate the quads. That is, we need to obtain the sequence of quads corresponding to the new DAG.

We need to construct a quad for every node that has a variable attached. If there are several variables attached we chose a live-on-exit variable, assuming we have done the necessary global flow analysis to determine such variables).

If there are several live-on-exit variables we need to compute one and make a copy so that we have both. An optimization pass may eliminate the copy if it is able to assure that one such variable may be used whenever the other is referenced.

Example

Recall the example from our movie

    a = b + c
    c = a + x
    d = b + c
    b = a + x
  

If b is dead on exit, the first three instructions suffice. If not we produce instead

    a = b + c
    c = a + x
    d = b + c
    b = c
  

If global analysis shows that, whenever this definition of b is used, c contains the same value, we can eliminate the copy and use c in place of b.

Order of Generated Instructions

Note that of the following 5 rules, 2 are due to arrays, and 2 due to pointers.

1. The DAG order must be respected (defs before uses).
2. Assignment to an array must follow all assignments to or uses of the same array that preceded it in the original block (no reordering of array assignments).
3. Uses of an array must follow all (preceding according to the original block) assignments to it; so the only transformation possible is reordering uses.
4. All variable references must follow all (preceding ...) procedure calls or assignment through a pointer.
5. A procedure call or assignment through a pointer must follow all (preceding ...) variable references.

Homework: 1, 2,

8.6: A Simple Code Generator

A big issue is proper use of the registers, which are often in short supply, and which are used/required for several purposes.

1. Some operands must be in registers.
2. Holding temporaries (i.e., values used in only one basic block) thereby avoiding expensive memory ops.
3. Holding inter-basic-block values (loop index).
4. Storage management (e.g., stack pointer).

For this section we assume a RISC architecture. Specifically, we assume only loads and stores touch memory; that is, the instruction set consists of

    LD  reg, mem
    ST  mem, reg
    OP  reg, reg, reg
  

A major simplification is we assume that, for each three address operation, there is precisely one machine instruction that accomplishes the task. This eliminates the question of instruction selection.

We do, however, consider register usage. Although we have not done global flow analysis (part of optimization), we will point out places where live-on-exit information would help us make better use of the available registers.

Recall that the mem operand in the load LD and store ST instructions can use any of the previously discussed addressing modes.

8.6.1: Register and Address Descriptors

These are the primary data structures used by the code generator. They keep track of what values are in each register as well as where a given value resides.

• Each register has a register descriptor containing the list of variables currently stored in this register. At the start of the basic block all register descriptors are empty.
• Each variable has a address descriptor containing the list of locations where this variable is currently stored. Possibilities are its memory location and one or more registers. The memory location might be in the static area, the stack, or presumably the heap (but not mentioned in the text).

The register descriptor could be omitted since you can compute it from the address descriptors.

8.6.2: The Code-Generation Algorithm

There are basically three parts to (this simple algorithm for) code generation.

1. Choosing registers
2. Generating instructions
3. Managing descriptors

We will isolate register allocation in a function getReg(Instruction), which is presented later. First presented is the algorithm to generate instructions. This algorithm uses getReg() and the descriptors. Then we learn how to manage the descriptors and finally we study getReg() itself.

Machine Instructions for Operations

Given a quad OP x, y, z (i.e., x = y OP z), proceed as follows.

1. Call getReg(OP x, y, z) to get R_x, R_y, and R_z, the registers to be used for x, y, and z respectively.

   Note that getReg merely selects the registers, it does not guarantee that the desired values are present in these registers.

2. Check the register descriptor for R_y. If y is not present in R_y, check the address descriptor for y and issue
         LD R_y, y'
   

   y' is some location containing y. Perhaps y is in a register other than R_y

3. Similar treatment for R_z.
   
   
4. Generate the instruction
         OP R_x, R_y, R_z
   Note that x now is not in its memory location.

Machine Instructions for Copy Statements

When processing
      x = y
steps 1 and 2 are analogous to the above, step 3 is vacuous, and step 4 is omitted since getReg() will set R_x=R_y.

Note that if y was already in a register before the copy instruction, no code is generated at this point (getReg will choose R_y to be a register containing y). Also note that since the value of x is now not in its memory location, we may need to store this value into x at block exit.

Ending the Basic Block

You may have noticed that we have not yet generated any store instructions. They occur here (and during spill code in getReg()). We need to ensure that all variables needed by (dynamically) subsequent blocks (i.e., those live-on-exit) have their current values in their memory locations.

1. Temporaries are never live beyond a basic block so can be ignored.
2. Variables dead on exit (thank you global flow analysis for determining such variables) are also ignored.
3. All live on exit variables (for all non-temporaries) need to be in their memory location on exit from the block. Therefore, for any live on exit variable whose own memory location is not listed in its address descriptor, generate ST x, R where R is a register listed in the address descriptor.

Managing Register and Address Descriptors

This is fairly clear. We just have to think through what happens when we do a load, a store, an (assembler) OP, or a copy. For R a register, let Desc(R) be its register descriptor. For x a program variable, let Desc(x) be its address descriptor.

1. Load: LD R, x
   • Desc(R) = x (removing everything else from Desc(R))
   • Add R to Desc(x) (leaving alone everything else in Desc(x))
   • Remove R from Desc(w) for all w ≠ x (not in 2e please check)
   
   
2. Store: ST x, R
   • Add the memory location of x to Desc(x)
   
   
3. Operation: OP R_x, R_y, R_z implementing the quad OP x, y, z
   • Desc(R_x) = x
   • Desc(x) = R_x (Now R_x does not contain x's memory location!)
   • Remove R_x from Desc(w) for all w ≠ x
   
   
4. Copy: For x = y after processing the load (if needed)
   • Add x to Desc(R_y) (recall that R_y=R_x).
   • Desc(x) = R_y.

Example

Since we haven't specified getReg() yet, we will assume there are an unlimited number of registers so we do not need to generate any spill code (saving the register's value in memory). One of getReg()'s jobs is to generate spill code when a register needs to be used for another purpose and the current value is not presently in memory.

Despite having ample registers and thus not generating spill code, we will not be wasteful of registers.

• When a register holds a temporary value and there are no subsequent uses of this value, we reuse that register.
• When a register holds the value of a program variable and there are no subsequent uses of this value, we reuse that register providing this value is also in the memory location for the variable.
• When a register holds the value of a program variable and all subsequent uses of this value are preceded by a redefinition, we could reuse this register. But to know about all subsequent uses may require live/dead-on-exit knowledge.

This example is from the book. I give another example after presenting getReg(), that I believe justifies my claim that the book is missing an action for load instructions, as indicated above.

Assume a, b, c, and d are program variables and t, u, v are compiler generated temporaries (I would call these t$1, t$2, and t$3). The intermediate language program is in the middle with the generated code for each quad shown. To the right is shown the contents of all the descriptors. The code generation is explained on the left.

t = a - b
    LD  R1, a
    LD  R2, b
    SUB R2, R1, R2


u = a - c
    LD  r3, c
    SUB R1, R1, R3


v = t + u
    ADD R3, R2, R1


a = d
    LD  R2, d


d = v + u
    ADD R1, R3, R1


exit
    ST  a, R2
    ST  d, R1

1. For the first quad, we need all three instructions since nothing is register resident on block entry. Since b is not used again, we can reuse its register. (Note that the current value of b is in its memory location.)
2. We do not load a again since its value is R1, which we can reuse for u since a is not used below.
3. We again reuse a register for the result; this time because c is not used again.
4. The copy instruction required a load since d was not in a register. As the descriptor shows, a was assigned to the same register, but no machine instruction was required.
5. The last instruction uses values already in registers. We can reuse R1 since u is a temporary.
6. At block exit, lacking global flow analysis, we must assume all program variables are live and hence must store back to memory any values located only in registers.

8.6.3: Design of the Function getReg

Consider
      x = y OP z
Picking registers for y and z are the same; we just do y. Choosing a register for x is a little different.

A copy instruction
      x = y
is easier.

Choosing R_y

Similar to demand paging, where the goal is to produce an available frame, our objective here is to produce an available register we can use for R_y. We apply the following steps in order until one succeeds. (Step 2 is a special case of step 3.)

1. If Desc(y) contains a register, use of these for R_y.
2. If Desc(R) is empty for some registers, pick one of these.
3. Pick a register for which the cleaning procedure generates a minimal number of store instructions. To clean an in-use register R do the following for each v in Desc(R).
   1. If Desc(v) includes something besides R, no store is needed for v.
   2. If v is x and x is not z, no store is needed since x is being overwritten.
   3. No store is needed if there is no further use of v prior to a redefinition. This is easy to check for further uses within the block. If v is live on exit (e.g., we have no global flow analysis), we need a redefinition later in this block.
   4. Otherwise a spill ST v, R is generated.

Choosing R_z and R_x, and Processing x = y

As stated above choosing R_z is the same as choosing R_y.

Choosing R_x has the following differences.

1. Since R_x will be written it is not enough for Desc(x) to contain a register R as in 1. above; instead, Desc(R) must contain only x.
2. If there is no further use of y prior to a redefinition (as described above for v) and if R_y contains only y (or will do so after it is loaded), then R_y can be used for R_x. Similarly, R_z might be usable for R_x.

getReg(x=y) chooses R_y as above and chooses R_x=R_y.

Example

                              R1  R2  R3    a    b    c    d    e
  			                  a    b    c    d    e

    a = b + c
        LD  R1, b
        LD  R2, c
        ADD R3, R1, R2
                              R1  R2  R3    a    b    c    d    e
  			    b   c   a     R3  b,R1 c,R2  d    e

    d = a + e
        LD  R1, e
        ADD R2, R3, R1
                              R1  R2  R3    a    b    c    d    e
                       2e →   e   d   a     R3  b,R1  c    R2  e,R1
                       me →   e   d   a     R3   b    c    R2  e,R1
  

We needed registers for d and e; none were free. getReg() first chose R2 for d since R2's current contents, the value of c, was also located in memory. getReg() then chose R1 for e for the same reason.

Using the 2e algorithm, b might appear to be in R1 (depends if you look in the address or register descriptors).

    a = e + d
        ADD R3, R1, R2
                              Descriptors unchanged

    e = a + b
        ADD R1, R3, R1   ← possible wrong answer from 2e
                              R1  R2  R3    a    b    c    d    e
                              e   d    a    R3  b,R1  c    R2   R1

        LD  R1, b
        ADD R1, R3, R1
                              R1  R2  R3    a    b    c    d    e
                              e   d    a    R3   b    c    R2   R1
  

The 2e might think R1 has b (address descriptor) and also conclude R1 has only e (register descriptor) so might generate the erroneous code shown.

Really b is not in a register so must be loaded. R3 has the value of a so was already chosen for a. R2 or R1 could be chosen. If R2 was chosen, we would need to spill d (we must assume live-on-exit, since we have no global flow analysis). We choose R1 since no spill is needed: the value of e (the current occupant of R1) is also in its memory location.

  exit
      ST  a, R3
      ST  d, R2
      ST  e, R1

8.7: Peephole Optimization

8.8: Register Allocation and Assignment

8.9: Instruction Selection by Tree Rewriting

What if a given quad needs several OPs and we have choices?

We would like to be able to describe the machine OPs in a way that enables us to find a sequence of OPs (and LDs and STs) to do the job.

The idea is that you express the quad as a tree and express each OP as a (sub-)tree simplification, i.e. the op replaces a subtree by a simpler subtree. In fact the simpler subtree is just a single node.

The diagram on the right represents x[i] = y[a] + 9, where x and y are on the stack and a is in the static area. M's are values in memory; C's are constants; and R's are registers. The weird ind (presumably short for indirect) treats its argument as a memory location.

Compare this to grammars: A production replaces the RHS by the LHS. We consider context free grammars where the LHS is a single nonterminal.

For example, a LD replaces a Memory node with a Register node.

Another example is that ADD R_i, R_i, R_j replaces a subtree consisting of a + with both children registers (i and j) with a Register node (i).

As you do the pattern matching and reductions (apply the productions), you emit the corresponding code (semantic actions). So to support a new processor, you need to supply the tree transformations corresponding to every instruction in the instruction set.

8.10: Optimal Code Generation for Expressions

This is quite cute.

We assume all operators are binary and label the instruction tree with something like the height. This gives the minimum number of registers needed so that no spill code is required. A few details follow.

8.10.1: Ershov Numbers

1. Draw the expression tree, the abstract syntax tree for an expression.
2. Label the leaves with 1.
3. Label interior nodes with L:
   1. If the children have the same label x, L=x+1. This looks like height.
   2. If the children have different labels, x and y, L=max(x,y).

8.10.2: Generating Code From Labeled Expression Trees

1. Recursive algorithm starting at the root. Each node puts its answer in the highest number register it is assigned. The idea is that a node uses (mostly) the same registers as its sibling.
   1. If the labels on the children are equal to L, the parent's label is L+1.
      1. Give one child L regs ending in highest assigned to parent. Note that the lowest reg assigned to the parent is not used by this child. The answer appears in top reg assigned to the child, which is the top reg assigned to the parent.
      2. Give other child L regs, ending one below the top reg assigned to the parent. This child does use the bottom reg assigned to the parent. The answer appears in top reg assigned to the child, i.e., the penultimate parent reg.
      3. Parent uses a two address OP to compute its answer in the same reg used by first child, which is the top reg assigned to the parent.
   2. If the labels on the children are M<L, the parent is labeled L.
      1. Give bigger child all L parent regs.
      2. Give other child M regs ending one below bigger child.
      3. Parent uses 2-addr OP computing answer in L
   3. If at a leaf (operand), load it into assigned reg.

Can see this is optimal (assuming you have enough registers).

1. Loads each operand only once.
2. Performs each operation only once.
3. Does no stores.
4. Minimal number of registers having the above three properties.
   1. Show need L registers to produce a result with label L.
   2. Must compute one side and not use the register containing its answer before finishing the other side.
   3. Apply this argument recursively.

8.10.3: Evaluating Expressions with an Insufficient Supply of Registers

Rough idea is to apply the above recursive algorithm, but at each recursive step, if the number of regs is not enough, store the result of the first child computed before starting the second.

8.11: Dynamic Programming Code-Generation