CS 406/534 Compiler Construction Parsing Part I


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1 CS 406/534 Compiler Construction Parsing Part I Prof. Li Xu Dept. of Computer Science UMass Lowell Fall 2004 Part of the course lecture notes are based on Prof. Keith Cooper, Prof. Ken Kennedy and Dr. Linda Torczon s teaching materials at Rice University. All rights reserved. 1
2 What We Did Last Time The cycle in lexical analysis RE NFA NFA DFA DFA Minimal DFA DFA RE Engineering issues in building scanners CS406/534 Fall 2004, Prof. Li Xu 2 2
3 Parsing Part I Today s Goals Contextfree grammars Sentence derivations Grammar ambiguity Left recursion problem with topdown parsing and how to fix it Predictive topdown parsing LL(1) condition Recursive descent parsing CS406/534 Fall 2004, Prof. Li Xu 3 3
4 Compilers Front Middle End Back End 4 Register Allocation Instruction Scheduling Instruction Selection IR Optimization n Optimization 2 Optimization 1 IR CSA Parser Scanner Infrastructure: symbol tables, trees, graphs, intermediate representations, sets, tuples CS406/534 Fall 2004, Prof. Li Xu 4 Analysis
5 The Front End Source code Scanner tokens Parser IR Parser Errors Checks the stream of words and their parts of speech (produced by the scanner) for grammatical correctness Determines if the input is syntactically well formed Guides checking at deeper levels than syntax Builds an IR representation of the code Think of this as the mathematics of diagramming sentences CS406/534 Fall 2004, Prof. Li Xu 5 5
6 The Study of Parsing The process of discovering a derivation for some sentence Need a mathematical model of syntax a grammar G Need an algorithm for testing membership in L(G) Need to keep in mind that our goal is building parsers, not studying the mathematics of arbitrary languages Roadmap 1 Contextfree grammars and derivations 2 Topdown parsing Handcoded recursive descent parsers LL(1) parsers LL(1) parsed topdown, left to right scan, leftmost derivation, 1 symbol lookahead 3 Bottomup parsing LR(1) parsers LR(1) parsed bottomup, left to right scan, reverse rightmost derivation, 1 symbol lookahead CS406/534 Fall 2004, Prof. Li Xu 6 6
7 Specifying Syntax with a Grammar Contextfree syntax is specified with a contextfree grammar SheepNoise SheepNoise baa baa This CFG defines the set of noises sheep normally make It is written in a variant of Backus Naur form Formally, a grammar is a four tuple, G = (S,N,T,P) S is the start symbol (set of strings in L(G)) N is a set of nonterminal symbols (syntactic variables) T is a set of terminal symbols (words) P is a set of productions or rewrite rules (P : N (N T) + ) CS406/534 Fall 2004, Prof. Li Xu 7 7
8 The Big Picture Chomsky Hierarchy of Language Grammars (1956) recursive enumerable context sensitive grammar CFG LR(1) LL(1) RE CS406/534 Fall 2004, Prof. Li Xu 8 8
9 Deriving Syntax We can use the SheepNoise grammar to create sentences use the productions as rewriting rules Rule Sentential Form SheepNoise 2 baa Rule Sentential Form SheepNoise 1 SheepNoise baa 2 baa baa Rule Sentential Form SheepNoise 1 SheepNoise baa 1 SheepNoise baa baa 2 baa baa baa And so on... While it is cute, this example quickly runs out of intellectual steam... CS406/534 Fall 2004, Prof. Li Xu 9 9
10 A More Useful Grammar To explore the uses of CFGs, we need a more complex grammar 1 Expr Expr Op Expr 2 number 3 id 4 Op * 7 / Rule Expr Sentential Form 1 Expr Op Expr 2 <id,x> Op Expr 5 <id,x> Expr 1 <id,x> Expr Op Expr 2 <id,x> <num,2> Op Expr 6 <id,x> <num,2> * Expr 3 <id,x> <num,2> * <id,y> Such a sequence of rewrites is called a derivation Process of discovering a derivation is called parsing We denote this derivation: Expr * id num* id CS406/534 Fall 2004, Prof. Li Xu 10 10
11 Derivations At each step, we choose a nonterminal to replace Different choices can lead to different derivations Two derivations are of interest Leftmost derivation replace leftmost NT at each step Rightmost derivation replace rightmost NT at each step These are the two systematic derivations (We don t care about randomlyordered derivations!) The example on the preceding slide was a leftmost derivation Of course, there is also a rightmost derivation Interestingly, it turns out to be different CS406/534 Fall 2004, Prof. Li Xu 11 11
12 The Two Derivations for x 2* y Rule Sentential Form Expr 1 Expr Op Expr 3 <id,x> Op Expr 5 <id,x> Expr 1 <id,x> Expr Op Expr 2 <id,x> <num,2> Op Expr 6 <id,x> <num,2> * Expr 3 <id,x> <num,2> * <id,y> Leftmost derivation Rule Sentential Form Expr 1 Expr Op Expr 3 Expr Op <id,y> 6 Expr * <id,y> 1 Expr Op Expr * <id,y> 2 Expr Op <num,2> * <id,y> 5 Expr <num,2> * <id,y> 3 <id,x> <num,2> * <id,y> Rightmost derivation In both cases, Expr * id num * id The two derivations produce different parse trees The parse trees imply different evaluation orders! CS406/534 Fall 2004, Prof. Li Xu 12 12
13 Derivations and Parse Trees Leftmost derivation Rule Sentential Form G Expr 1 Expr Op Expr 3 <id,x> Op Expr E 5 <id,x> Expr 1 <id,x> Expr Op Expr 2 <id,x> <num,2> Op Expr E Op E 6 <id,x> <num,2> * Expr 3 <id,x> <num,2> * <id,y> x E Op E This evaluates as x ( 2* y ) 2 * y CS406/534 Fall 2004, Prof. Li Xu 13 13
14 Derivations and Parse Trees Rightmost derivation Rule Sentential Form Expr 1 Expr Op Expr 3 Expr Op <id,y> 6 Expr * <id,y> 1 Expr Op Expr * <id,y> 2 Expr Op <num,2> * <id,y> 5 Expr <num,2> * <id,y> 3 <id,x> <num,2> * <id,y> E G E Op E E Op E * y This evaluates as ( x 2 ) * y x 2 CS406/534 Fall 2004, Prof. Li Xu 14 14
15 Derivations and Precedence These two derivations point out a problem with the grammar: It has no notion of precedence, or implied order of evaluation To add precedence Create a nonterminal for each level of precedence Isolate the corresponding part of the grammar Force the parser to recognize high precedence subexpressions first For algebraic expressions Multiplication and division, first (level one) Subtraction and addition, next (level two) CS406/534 Fall 2004, Prof. Li Xu 15 15
16 Derivations and Precedence Adding the standard algebraic precedence produces: level two level one 1 Goal Expr 2 Expr Expr + Term 3 Expr Term 4 Term 5 Term Term * Factor 6 Term / Factor 7 Factor 8 Factor number 9 id This grammar is slightly larger Takes more rewriting to reach some of the terminal symbols Encodes expected precedence Produces same parse tree under leftmost & rightmost derivations Let s see how it parses x  2 * y CS406/534 Fall 2004, Prof. Li Xu 16 16
17 Derivations and Precedence Rule Sentential Form Goal G 1 Expr E 3 Expr Term 5 Expr Term * Factor E T 9 Expr Term * <id,y> 7 Expr Factor * <id,y> T T * F 8 Expr <num,2> * <id,y> 4 Term <num,2> * <id,y> F F <id,y> 7 Factor <num,2> * <id,y> <id,x> <num,2> 9 <id,x> <num,2> * <id,y> This produces x ( 2* y ), along with an appropriate parse tree. Both the leftmost and rightmost derivations give the same expression, because the grammar directly encodes the desired precedence. CS406/534 Fall 2004, Prof. Li Xu 17 17
18 Ambiguous Grammars Our original expression grammar had other problems 1 Expr Expr Op Expr 2 number 3 id 4 Op * 7 / Rule Sentential Form Expr 1 Expr Op Expr 1 Expr Op Expr Op Expr 3 <id,x> Op Expr Op Expr 5 <id,x> Expr Op Expr 2 <id,x> <num,2> Op Expr 6 <id,x> <num,2> * Expr 3 <id,x> <num,2> * <id,y> This grammar allows multiple leftmost derivations for x 2* y Hard to automate derivation if > 1 choice The grammar is ambiguous different choice than the first time CS406/534 Fall 2004, Prof. Li Xu 18 18
19 Two Leftmost Derivations for x 2 * y The Difference: Different productions chosen on the second step Rule Sentential Form Expr 1 Expr Op Expr 3 <id,x> Op Expr 5 <id,x> Expr 1 <id,x> Expr Op Expr 2 <id,x> <num,2> Op Expr 6 <id,x> <num,2> * Expr 3 <id,x> <num,2> * <id,y> Original choice Both derivations succeed in producing x  2 * y Rule Sentential Form Expr 1 Expr Op Expr 1 Expr Op Expr Op Expr 3 <id,x> Op Expr Op Expr 5 <id,x> Expr Op Expr 2 <id,x> <num,2> Op Expr 6 <id,x> <num,2> * Expr 3 <id,x> <num,2> * <id,y> New choice CS406/534 Fall 2004, Prof. Li Xu 19 19
20 Definitions Ambiguous Grammars If a grammar has more than one leftmost derivation for a single sentential form, the grammar is ambiguous If a grammar has more than one rightmost derivation for a single sentential form, the grammar is ambiguous The leftmost and rightmost derivations for a sentential form may differ, even in an unambiguous grammar Classic example the ifthenelse problem Stmt if Expr then Stmt if Expr then Stmt else Stmt other stmts This ambiguity is entirely grammatical in nature CS406/534 Fall 2004, Prof. Li Xu 20 20
21 Ambiguity This sentential form has two derivations if Expr 1 then if Expr 2 then Stmt 1 else Stmt 2 if if E 1 then else E 1 then if S 2 if E 2 then E 2 then else S 1 S 1 S 2 production 2, then production 1 production 1, then production 2 CS406/534 Fall 2004, Prof. Li Xu 21 21
22 Removing the ambiguity Ambiguity Must rewrite the grammar to avoid generating the problem Match each else to innermost unmatched if (common sense rule) 1 Stmt WithElse 2 NoElse 3 WithElse if Expr then WithElse else WithElse 4 OtherStmt 5 NoElse if Expr then Stmt 6 if Expr then WithElse else NoElse Intuition: a NoElse always has no else on its last cascaded else if statement With this grammar, the example has only one derivation CS406/534 Fall 2004, Prof. Li Xu 22 22
23 Ambiguity if Expr 1 then if Expr 2 then Stmt 1 else Stmt 2 Rule Sentential Form Stmt 2 NoElse 5 if Expr then Stmt? if E 1 then Stmt 1 if E 1 then WithElse 3 if E 1 then if Expr then WithElse else WithElse? if E 1 then if E 2 then WithElse else WithElse 4 if E 1 then if E 2 then S 1 else WithElse 4 if E 1 then if E 2 then S 1 else S 2 This binds the else controlling S 2 to the inner if CS406/534 Fall 2004, Prof. Li Xu 23 23
24 Deeper Ambiguity Ambiguity usually refers to confusion in the CFG Overloading can create deeper ambiguity a = f(17) In many Algollike languages, f could be either a function or a subscripted variable Disambiguating this one requires context Need values of declarations Really an issue of type, not contextfree syntax Requires an extragrammatical solution (not in CFG) Must handle these with a different mechanism Step outside grammar rather than use a more complex grammar CS406/534 Fall 2004, Prof. Li Xu 24 24
25 Ambiguity  The Final Word Ambiguity arises from two distinct sources Confusion in the contextfree syntax (ifthenelse) Confusion that requires context to resolve (overloading) Resolving ambiguity To remove contextfree ambiguity, rewrite the grammar To handle contextsensitive ambiguity takes cooperation Knowledge of declarations, types, Accept a superset of L(G) & check it by other means This is a language design problem Sometimes, the compiler writer accepts an ambiguous grammar Parsing techniques that do the right thing i.e., always select the same derivation CS406/534 Fall 2004, Prof. Li Xu 25 25
26 Parsing Techniques Topdown parsers (LL(1), recursive descent) Start at the root of the parse tree and grow toward leaves Pick a production & try to match the input Bad pick may need to backtrack Some grammars are backtrackfree (predictive parsing) Bottomup parsers (LR(1), operator precedence) Start at the leaves and grow toward root As input is consumed, encode possibilities in an internal state Start in a state valid for legal first tokens Bottomup parsers handle a large class of grammars CS406/534 Fall 2004, Prof. Li Xu 26 26
27 Topdown Parsing A topdown parser starts with the root of the parse tree The root node is labeled with the goal symbol of the grammar Topdown parsing algorithm: Construct the root node of the parse tree Repeat until the fringe of the parse tree matches the input string 1 At a node labeled A, select a production with A on its lhs and, for each symbol on its rhs, construct the appropriate child 2 When a terminal symbol is added to the fringe and it doesn t match the fringe, backtrack 3 Find the next node to be expanded (label NT) The key is picking the right production in step 1 That choice should be guided by the input string CS406/534 Fall 2004, Prof. Li Xu 27 27
28 The Expression Grammar Version with precedence derived last lecture 1 Goal Expr 2 Expr Expr + Term 3 Expr Term 4 Term 5 Term Term * Factor 6 Term / Factor 7 Factor 8 Factor number 9 id And the input x 2* y CS406/534 Fall 2004, Prof. Li Xu 28 28
29 Example Let s try x 2* y : Rule Sentential Form Input Goal Goal x 2 * y Expr 1 Expr x 2 * y 2 Expr + Term x 2 * y Expr + Term 4 Term + Term x 2 * y Term 7 Factor + Term x 2 * y 9 <id,x> + Term x 2 * y Fact. 9 <id,x> + Term x 2 * y <id,x> Leftmost derivation, choose productions in an order that exposes problems CS406/534 Fall 2004, Prof. Li Xu 29 29
30 Let s try x 2* y : Example Goal Rule Sentential Form Input Goal x 2 * y Expr 1 Expr x 2 * y Expr + Term 2 Expr + Term x 2 * y 4 Term + Term x 2 * y Term 7 Factor + Term x 2 * y 9 <id,x> + Term x 2 * y Fact. 9 <id,x> + Term x 2 * y <id,x> This worked well, except that doesn t match + The parser must backtrack to here CS406/534 Fall 2004, Prof. Li Xu 30 30
31 Continuing with x 2* y : Example Goal Rule Sentential Form Input Goal x 2 * y Expr 1 Expr x 2 * y 3 Expr Term x 2 * y Expr Term 4 Term Term x 2 * y 7 Factor Term x 2 * y Term 9 <id,x> Term x 2 * y 9 <id,x> Term x 2 * y <id,x> Term x 2 * y Fact. <id,x> CS406/534 Fall 2004, Prof. Li Xu 31 31
32 Example Continuing with x 2* y : Goal Rule Sentential Form Input Goal x 2 * y Expr 1 Expr x 2 * y 3 Expr Term x 2 * y Expr Term 4 Term Term x 2 * y 7 Factor Term x 2 * y Term 9 <id,x> Term x 2 * y 9 <id,x> Term x 2 * y <id,x> Term x 2 * y Fact. <id,x> This time, and matched We can advance past to look at 2 Now, we need to expand Term the last NT on the fringe CS406/534 Fall 2004, Prof. Li Xu 32 32
33 Example Trying to match the 2 in x 2* y : Goal Rule Sentential Form Input <id,x> Term x 2 * y Expr 7 <id,x> Factor x 2 * y Expr Term 9 <id,x> <num,2> x 2 * y <id,x> <num,2> x 2 * y Term Fact. Fact. <num,2> <id,x> CS406/534 Fall 2004, Prof. Li Xu 33 33
34 Example Trying to match the 2 in x 2* y : Goal Rule Sentential Form Input <id,x> Term x 2 * y Expr 7 <id,x> Factor x 2 * y Expr  Term 9 <id,x> <num,2> x 2 * y <id,x> <num,2> x 2 * y Term Fact. Where are we? 2 matches 2 Fact. <id,x> We have more input, but no NTs left to expand The expansion terminated too soon Need to backtrack <num,2> CS406/534 Fall 2004, Prof. Li Xu 34 34
35 Example Trying again with 2 in x 2* y : Goal Rule Sentential Form Input <id,x> Term x 2 * y Expr 5 <id,x> Term * Factor x 2 * y Expr Term 7 <id,x> Factor * Factor x 2 * y 8 <id,x> <num,2> * Factor x 2 * y Term Term * Fact. <id,x> <num,2> * Factor x 2 * y <id,x> <num,2> * Factor x 2 * y Fact. Fact. <id,y> 9 <id,x> <num,2> * <id,y> x 2 * y <id,x> <num,2> <id,x> <num,2> * <id,y> x 2 * y This time, we matched & consumed all the input Success! CS406/534 Fall 2004, Prof. Li Xu 35 35
36 Another Possible Parse Other choices for expansion are possible Rule Sentential Form Input Goal x 2 * y 1 Expr x 2 * y 2 Expr + Term x 2 * y 2 Expr + Term +Term x 2 * y 2 Expr + Term + Term +Term x 2 * y 2 Expr +Term + Term + +Term x 2 * y consuming no input! This doesn t terminate (obviously) Wrong choice of expansion leads to nontermination Nontermination is a bad property for a parser to have Parser must make the right choice CS406/534 Fall 2004, Prof. Li Xu 36 36
37 Left Recursion Topdown parsers cannot handle leftrecursive grammars Formally, A grammar is left recursive if A NT such that a derivation A + Aα, for some string α (NT T ) + Our expression grammar is left recursive This can lead to nontermination in a topdown parser For a topdown parser, any recursion must be right recursion We would like to convert the left recursion to right recursion Nontermination is a bad property in any part of a compiler CS406/534 Fall 2004, Prof. Li Xu 37 37
38 Eliminating Left Recursion To remove left recursion, we can transform the grammar Consider a grammar fragment of the form Fee Fee α β where neither α nor β start with Fee We can rewrite this as Fee β Fie Fie α Fie ε where Fie is a new nonterminal This accepts the same language, but uses only right recursion CS406/534 Fall 2004, Prof. Li Xu 38 38
39 Eliminating Left Recursion The expression grammar contains two cases of left recursion Expr Expr + Term Term Term * Factor Expr Term Term / Factor Term Factor Applying the transformation yields Expr Term Expr Expr + Term Expr Term Expr ε Term Factor Term Term * Factor Term / Factor Term ε These fragments use only right recursion They retain the original left associativity CS406/534 Fall 2004, Prof. Li Xu 39 39
40 Eliminating Left Recursion Substituting them back into the grammar yields 1 Goal Expr 2 Expr Term Expr 3 Expr + Term Expr 4 Term Expr 5 ε 6 Term Factor Term 7 Term * Factor Term 8 / Factor Term 9 ε 10 Factor number 11 id 12 ( Expr ) This grammar is correct, if somewhat nonintuitive. It is left associative, as was the original A topdown parser will terminate using it. A topdown parser may need to backtrack with it. CS406/534 Fall 2004, Prof. Li Xu 40 40
41 Eliminating Left Recursion The transformation eliminates immediate left recursion What about more general, indirect left recursion? The general algorithm: arrange the NTs into some order A 1, A 2,, A n for i 1 to n for s 1 to i 1 Must start with 1 to ensure that A 1 A 1 β is transformed replace each production A i A s γ with A i δ 1 γ δ 2 γ δ k γ, where A s δ 1 δ 2 δ k are all the current productions for A s eliminate any immediate left recursion on A i using the direct transformation This assumes that the initial grammar has no cycles (A i + A i ), and no epsilon productions And back CS406/534 Fall 2004, Prof. Li Xu 41 41
42 Eliminating Left Recursion How does this algorithm work? 1. Impose arbitrary order on the nonterminals 2. Outer loop cycles through NT in order 3. Inner loop ensures that a production expanding A i has no nonterminal A s in its rhs, for s < i 4. Last step in outer loop converts any direct recursion on A i to right recursion using the transformation showed earlier 5. New nonterminals are added at the end of the order & have no left recursion At the start of the i th outer loop iteration For all k < i, no production that expands A k contains a nonterminal A s in its rhs, for s < k CS406/534 Fall 2004, Prof. Li Xu 42 42
43 Example Order of symbols: G, E, T G E E E + T E T T E ~ T T id CS406/534 Fall 2004, Prof. Li Xu 43 43
44 Example Order of symbols: G, E, T 1. A i = G G E E E + T E T T E ~ T T id CS406/534 Fall 2004, Prof. Li Xu 44 44
45 Example Order of symbols: G, E, T 1. A i = G G E E E + T E T T E ~ T T id 2. A i = E G E E T E' E' + T E' E' ε T E ~ T T id CS406/534 Fall 2004, Prof. Li Xu 45 45
46 Example Order of symbols: G, E, T 1. A i = G G E E E + T E T T E ~ T T id 2. A i = E G E E T E' E' + T E' E' ε T E ~ T T id 3. A i = T, A s = E G E E T E' E' + T E' E' ε T T E' ~ T T id Go to Algorithm CS406/534 Fall 2004, Prof. Li Xu 46 46
47 Example Order of symbols: G, E, T 1. A i = G 2. A i = E 3. A i = T, A s = E 4. A i = T G E G E G E G E E E + T E T E' E T E' E T E' E T E' + T E' E' + T E' E' + T E' T E ~ T E' ε E' ε E' ε T id T E ~ T T T E' ~ T T id T' T id T id T' E' ~ T T' T' ε CS406/534 Fall 2004, Prof. Li Xu 47 47
48 Roadmap (Where are We?) We set out to study parsing Specifying syntax Contextfree grammars Ambiguity Topdown parsers Algorithm & its problem with left recursion Leftrecursion removal Predictive topdown parsing The LL(1) condition Simple recursive descent parsers CS406/534 Fall 2004, Prof. Li Xu 48 48
49 Picking the Right Production If it picks the wrong production, a topdown parser may backtrack Alternative is to look ahead in input & use context to pick correctly How much lookahead is needed? In general, an arbitrarily large amount Use the CockeYounger, Kasami algorithm or Earley s algorithm Fortunately, Large subclasses of CFGs can be parsed with limited lookahead Most programming language constructs fall in those subclasses Among the interesting subclasses are LL(1) grammars and LR(1) CS406/534 Fall 2004, Prof. Li Xu 49 49
50 Basic idea Predictive Parsing Given A α β, the parser should be able to choose between α & β FIRST sets For some rhs α G, define FIRST(α) as the set of tokens that appear as the first symbol in some string that derives from α That is, x FIRST(α) iff α * x γ, for some γ We will defer the problem of how to compute FIRST sets until we look at the LR(1) table construction algorithm CS406/534 Fall 2004, Prof. Li Xu 50 50
51 Predictive Parsing Basic idea Given A α β, the parser should be able to choose between α & β FIRST sets For some rhs α G, define FIRST(α) as the set of tokens that appear as the first symbol in some string that derives from α That is, x FIRST(α) iff α * x γ, for some γ The LL(1) Property If A αand A βboth appear in the grammar, we would like FIRST(α) FIRST(β) = This would allow the parser to make a correct choice with a lookahead of exactly one symbol! This is almost correct See the next slide CS406/534 Fall 2004, Prof. Li Xu 51 51
52 Predictive Parsing What about εproductions? They complicate the definition of LL(1) If A αand A βand ε FIRST(α), then we need to ensure that FIRST(β) is disjoint from FOLLOW(α), too Define FIRST + (α) as FIRST(α) FOLLOW(α), if ε FIRST(α) FIRST(α), otherwise Then, a grammar is LL(1) iff A αand A βimplies FIRST + (α) FIRST + (β) = FOLLOW(α) is the set of all words in the grammar that can legally appear immediately after an α CS406/534 Fall 2004, Prof. Li Xu 52 52
53 Predictive Parsing Given a grammar that has the LL(1) property Can write a simple routine to recognize each lhs Code is both simple & fast Consider A β 1 β 2 β 3, with FIRST + (β 1 ) FIRST + (β 2 ) FIRST + (β 3 ) = /* find an A */ if (current_word FIRST(β 1 )) find a β 1 and return true else if (current_word FIRST(β 2 )) find a β 2 and return true else if (current_word FIRST(β 3 )) find a β 3 and return true else report an error and return false Of course, there is more detail to find a β i ( in EAC) Grammars with the LL(1) property are called predictive grammars because the parser can predict the correct expansion at each point in the parse. Parsers that capitalize on the LL(1) property are called predictive parsers. One kind of predictive parser is the recursive descent parser. CS406/534 Fall 2004, Prof. Li Xu 53 53
54 Recursive Descent Parsing Recall the expression grammar, after transformation 1 Goal Expr 2 Expr Term Expr 3 Expr + Term Expr 4 Term Expr 5 ε 6 Term Factor Term 7 Term * Factor Term 8 / Factor Term 9 ε 10 Factor number 11 id This produces a parser with six mutually recursive routines: Goal Expr EPrime Term TPrime Factor Each recognizes one NT or T The term descent refers to the direction in which the parse tree is built. CS406/534 Fall 2004, Prof. Li Xu 54 54
55 Recursive Descent Parsing A couple of routines from the expression parser Goal( ) token next_token( ); if (Expr( ) = true & token = EOF) then next compilation step; else report syntax error; return false; Expr( ) if (Term( ) = false) then return false; else return Eprime( ); looking for EOF, found token Factor( ) if (token = Number) then token next_token( ); return true; else if (token = Identifier) then token next_token( ); return true; else report syntax error; return false; EPrime, Term, & TPrime follow the same basic lines (Figure 3.7, EAC) looking for Number or Identifier, found token instead CS406/534 Fall 2004, Prof. Li Xu 55 55
56 Recursive Descent Parsing To build a parse tree: Augment parsing routines to build nodes Pass nodes between routines using a stack Node for each symbol on rhs Action is to pop rhs nodes, make them children of lhs node, and push this subtree To build an abstract syntax tree Build fewer nodes Put them together in a different order Expr( ) result true; if (Term( ) = false) then return false; else if (EPrime( ) = false) then result false; else build an Expr node pop EPrime node pop Term node make EPrime & Term children of Expr push Expr node return result; Success build a piece of the parse tree CS406/534 Fall 2004, Prof. Li Xu 56 56
57 Left Factoring What if my grammar does not have the LL(1) property? Sometimes, we can transform the grammar The Algorithm A NT, find the longest prefix α that occurs in two or more righthand sides of A if α ε then replace all of the A productions, A αβ 1 αβ 2 αβ n γ, with A α Z γ Z β 1 β 2 β n where Z is a new element of NT Repeat until no common prefixes remain CS406/534 Fall 2004, Prof. Li Xu 57 57
58 Left Factoring A graphical explanation for the same idea αβ 1 A αβ 1 αβ 2 αβ3 A αβ 2 becomes αβ 3 A α Z Z β 1 β 2 β n A αz β 1 β 2 β 3 CS406/534 Fall 2004, Prof. Li Xu 58 58
59 Left Factoring: An Example Consider the following fragment of the expression grammar Factor Identifier Identifier [ ExprList ] Identifier ( ExprList ) FIRST(rhs 1 ) = { Identifier } FIRST(rhs 2 ) = { Identifier } FIRST(rhs 3 ) = { Identifier } After left factoring, it becomes Factor Identifier Arguments Arguments [ ExprList ] ( ExprList ) ε FIRST(rhs 1 ) = { Identifier } FIRST(rhs 2 ) = { [ } FIRST(rhs 3 ) = { ( } FIRST(rhs 4 ) = FOLLOW(Factor) It has the LL(1) property This form has the same syntax, with the LL(1) property CS406/534 Fall 2004, Prof. Li Xu 59 59
60 Graphically Left Factoring Identifier Factor Identifier [ ExprList ] No basis for choice Identifier ( ExprList ) becomes ε Factor Identifier [ ExprList ] Word determines correct choice ( ExprList ) CS406/534 Fall 2004, Prof. Li Xu 60 60
61 Recursive Descent (Summary) 1. Build FIRST (and FOLLOW) sets 2. Massage grammar to have LL(1) condition a. Remove left recursion b. Left factor it 3. Define a procedure for each nonterminal a. Implement a case for each righthand side b. Call procedures as needed for nonterminals 4. Add extra code, as needed a. Perform contextsensitive checking b. Build an IR to record the code Can we automate this process? CS406/534 Fall 2004, Prof. Li Xu 61 61
62 Summary Parsing Part I Introduction to parsing grammar, derivation, ambiguity, left recursion Predictive topdown parsing LL(1) condition Recursive descent parsing CS406/534 Fall 2004, Prof. Li Xu 62 62
63 Next Class Tabledriven LL(1) parsing Bottomup parsing CS406/534 Fall 2004, Prof. Li Xu 63 63
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