Compilers

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.

As I have mentioned before, when the entire grammar is written, no conventions are needed to tell the non-terminals, terminals, and start symbol. The non-terminals are the LHSs, the terminals are everything else on the RHS, and the start symbol is the LHS of the first production.

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 string—without looking back in the notes—with two derivations from E that give 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.

Definition: If x can be derived using a leftmost derivation, we call x a left-sentential form. Similarly for a right-sentential form.

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 non-terminal 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 non-terminal 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
  

    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
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 non-terminals 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 → a 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 a'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 is 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 *. With the recursive grammar, we have the following lm derivation.
A ⇒ A + ⇒ A + + ⇒ * + +
With the non-recursive grammar we have
A ⇒ * A' ⇒ * + A' ⇒ * + + A' ⇒ * + +

This 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 → x y1 | x y2 into

   A  → x A'
   A' → y1 | y2
  

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 non-terminal 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 processing.
   
   
2. Choose a production having the current nonterminal A as LHS. Say the RHS is X1 X2 ... Xn.
   
   

3. for i = 1 to n
     if Xi is a nonterminal
       process Xi  // recursive
     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. Since this is recursive, it is possible that no productions work for this nonterminal, because the wrong choice was made earlier.

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.

1. FIRST(a)={a} for all terminals a.
2. Initialize FIRST(A)=φ for all nonterminals A
3. If A → ε is a production, add ε to FIRST(A).
4. 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 this entire step until nothing is added.
5. FIRST of any string X=X₁X₂...X_n is initialized to φ and then
   1. add to FIRST(X) any non-ε symbol in FIRST(X_i) if ε is in all previous FIRST(X_j).
   2. add ε to FIRST(X) if ε is in every FIRST(X_j). In particular if X is ε, FIRST(X)={ε}.
6. Initialize FOLLOW(S)=$ and FOLLOW(A)=φ for all other nonterminals A, and then apply the following three rules until nothing is added to any FOLLOW set.
   1. For every production A → α B β, add all of FIRST(β) except ε to FOLLOW(B).
   2. For every production A → α B, add all of FOLLOW(A) to FOLLOW(B).
   3. For every production 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