Compilers

Start Lecture #4

Remark: Lab 2 assigned. It is due in 2 weeks, 27 September 2008.

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 - 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.

The reason we don't include the positive closure is that for any RE
r⁺ = rr^*.

Homework: 2(a-d), 4.

Examples
Let the alphabet be {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)!

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)?
         

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 automata (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. Regular expressions are a key in this task. So we want to recognize regular expressions (especially the ones representing tokens). 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 a DFA D that accepts the same strings as N does. 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 this 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.

NFA states	DFA state	a	b
{0,1,2,4,7}	D0	D1	D2
{1,2,3,4,6,7,8}	D1	D1	D3
{1,2,4,5,6,7}	D2	D1	D2
{1,2,4,5,6,7,9}	D3	D1	D4
{1,2,4,5,6,7,10}	D4	D1	D2

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 is 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.

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 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

3.7.5: Efficiency of String-Processing Algorithms

Skipped.

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).

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.

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 diagram 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.

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 us trouble since it is left-recursive and we did top-down parsing. So we use 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