Lexical and Syntax Analysis


 Lorraine Little
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1 Lexical and Syntax Analysis (of Programming Languages) TopDown Parsing
2 Lexical and Syntax Analysis (of Programming Languages) TopDown Parsing
3 Easy for humans to write and understand String of characters Lexemes identified String of tokens Easy for programs to transform Data structure
4 Easy for humans to write and understand String of characters Lexemes identified String of tokens Easy for programs to transform Data structure
5 PART 1: SYNTAX OF LANGUAGES ContextFree Grammars Derivations Parse Trees Ambiguity Precedence and Associativity
6 PART 1: SYNTAX OF LANGUAGES ContextFree Grammars Derivations Parse Trees Ambiguity Precedence and Associativity
7 Syntax The syntax is a set of rules defining valid strings of a language, often specified by a contextfree grammar. For example, a grammar E for arithmetic expressions: e x y e + e e e e * e ( e )
8 Syntax The syntax is a set of rules defining valid strings of a language, often specified by a contextfree grammar. For example, a grammar E for arithmetic expressions: e x y e + e e e e * e ( e )
9 Contextfree grammars Have four components: 1. A set of terminal symbols. 2. A set of nonterminal symbols. 3. A set of productions (or rules) of the form: n X 1 X n where n is a nonterminal and X 1 X n is any sequence of terminals, nonterminals, and ε. 4. The start symbol (one of the nonterminals).
10 Contextfree grammars Have four components: 1. A set of terminal symbols. 2. A set of nonterminal symbols. 3. A set of productions (or rules) of the form: n X 1 X n where n is a nonterminal and X 1 X n is any sequence of terminals, nonterminals, and ε. 4. The start symbol (one of the nonterminals).
11 Notation Nonterminals are underlined. Rather than writing e x e e + e we may write: e x e + e (Also, symbols and ::= will be used interchangeably.)
12 Notation Nonterminals are underlined. Rather than writing e x e e + e we may write: e x e + e (Also, symbols and ::= will be used interchangeably.)
13 Why contextfree? Unrestricted Context Sensitive Context Free Regular Nice balance between expressive power and efficiency of parsing.
14 Why contextfree? Unrestricted Context Sensitive Context Free Regular Nice balance between expressive power and efficiency of parsing.
15 Derivations A derivation is a proof that the some string conforms to a grammar. For example: e e + e x + e x + ( e ) x + ( e * e ) x + ( y * e ) x + ( y * x )
16 Derivations A derivation is a proof that the some string conforms to a grammar. For example: e e + e x + e x + ( e ) x + ( e * e ) x + ( y * e ) x + ( y * x )
17 Derivations Leftmost derivation: always expand the leftmost nonterminal when applying the grammar rules. Rightmost derivation: always expand the rightmost nonterminal, e.g. e e + e e + ( e ) e + ( x ) x + ( x )
18 Derivations Leftmost derivation: always expand the leftmost nonterminal when applying the grammar rules. Rightmost derivation: always expand the rightmost nonterminal, e.g. e e + e e + ( e ) e + ( x ) x + ( x )
19 Parse tree: motivation Like a derivation: a proof that a given input is valid according to the grammar. But a parse tree: is more concise: we don t write out the sentence every time a nonterminal is expanded. abstracts over the order in which rules are applied.
20 Parse tree: motivation Like a derivation: a proof that a given input is valid according to the grammar. But a parse tree: is more concise: we don t write out the sentence every time a nonterminal is expanded. abstracts over the order in which rules are applied.
21 Parse tree: intuition If nonterminal n has a production n X Y Z where X, Y, and Z are terminals or nonterminals, then a parse tree may have an interior node labelled n with three children labelled X, Y, and Z. n X Y Z
22 Parse tree: intuition If nonterminal n has a production n X Y Z where X, Y, and Z are terminals or nonterminals, then a parse tree may have an interior node labelled n with three children labelled X, Y, and Z. n X Y Z
23 Parse tree: definition A parse tree is a tree in which: the root is labelled by the start symbol; each leaf is labelled by a terminal symbol, or ε; each interior node is labelled by a nonterminal; if n is a nonterminal labelling an interior node whose children are X 1, X 2,, X n then there must exist a production n X 1 X 2 X n.
24 Parse tree: definition A parse tree is a tree in which: the root is labelled by the start symbol; each leaf is labelled by a terminal symbol, or ε; each interior node is labelled by a nonterminal; if n is a nonterminal labelling an interior node whose children are X 1, X 2,, X n then there must exist a production n X 1 X 2 X n.
25 Example 1 Example input string: x + y * x Resulting parse tree according to grammar E: e e + e x e * e y x
26 Example 1 Example input string: x + y * x Resulting parse tree according to grammar E: e e + e x e * e y x
27 Example 2 The following is not a parse tree according to grammar E. e x + e e y * e x Why? Because e x + e is not a production in grammar E.
28 Example 2 The following is not a parse tree according to grammar E. e x + e e y * e x Why? Because e x + e is not a production in grammar E.
29 Syntax Analysis String of symbols Parse tree A parse tree is: 1. A proof that a given input is valid according to the grammar; 2. A structurerich representation of the input that can be stored in a data structure that is convenient to process. (Syntax analysis may also report that the input string is invalid.)
30 Syntax Analysis String of symbols Parse tree A parse tree is: 1. A proof that a given input is valid according to the grammar; 2. A structurerich representation of the input that can be stored in a data structure that is convenient to process. (Syntax analysis may also report that the input string is invalid.)
31 Ambiguity If there exists more than one parse tree for any string then the grammar is ambiguous. For example, the string x+y*x has two parse trees: e e e + e e * e x e * e e + e x y x x y
32 Ambiguity If there exists more than one parse tree for any string then the grammar is ambiguous. For example, the string x+y*x has two parse trees: e e e + e e * e x e * e e + e x y x x y
33 Operator precedence Different parse trees often have different meanings, so we usually want unambiguous grammars. Conventionally, * has a higher precedence (binds tighter) than +, so there is only one interpretation of x+y*x, namely x+(y*x).
34 Operator precedence Different parse trees often have different meanings, so we usually want unambiguous grammars. Conventionally, * has a higher precedence (binds tighter) than +, so there is only one interpretation of x+y*x, namely x+(y*x).
35 Operator associativity Even with operator precedence rules, ambiguity remains, e.g. xxx. Binary operators are either: leftassociative; rightassociative; nonassociative. Conventionally,  is leftassociative, so there is only one interpretation of xxx, namely (xx)x.
36 Operator associativity Even with operator precedence rules, ambiguity remains, e.g. xxx. Binary operators are either: leftassociative; rightassociative; nonassociative. Conventionally,  is leftassociative, so there is only one interpretation of xxx, namely (xx)x.
37 Exercise 1 Recall grammar E: e x y e + e e e e * e ( e ) Let all operators be left associative, and let * bind tighter than + and. Give an unambiguous grammar for expressions, using these rules of associativity and precedence.
38 Exercise 1 Recall grammar E: e x y e + e e e e * e ( e ) Let all operators be left associative, and let * bind tighter than + and. Give an unambiguous grammar for expressions, using these rules of associativity and precedence.
39 Answer: stepbystep Given a nonterminal e which involves operators at n levels of precedence: Step 1: introduce n+1 new nonterminals, e 0 e n.
40 Answer: stepbystep Given a nonterminal e which involves operators at n levels of precedence: Step 1: introduce n+1 new nonterminals, e 0 e n.
41 Let op denote an operator with precedence i. Step 2: replace each production with e e op e e i e i op e i+1 e i+1 if op is leftassociative, or e i e i+1 op e i e i+1 if op is rightassociative
42 Let op denote an operator with precedence i. Step 2: replace each production with e e op e e i e i op e i+1 e i+1 if op is leftassociative, or e i e i+1 op e i e i+1 if op is rightassociative
43 Construct the precedence table: Operator Precedence +,  0 * 1 Grammar E after step 2 becomes: e 0 e 0 + e 1 e 0 e 1 e 1 e 1 e 1 * e 2 e 2 e ( e ) x y
44 Construct the precedence table: Operator Precedence +,  0 * 1 Grammar E after step 2 becomes: e 0 e 0 + e 1 e 0 e 1 e 1 e 1 e 1 * e 2 e 2 e ( e ) x y
45 Step 3: replace each production with e e n After step 3: e 0 e 0 + e 1 e 0 e 1 e 1 e 1 e 1 * e 2 e 2 e 2 ( e ) x y
46 Step 3: replace each production with e e n After step 3: e 0 e 0 + e 1 e 0 e 1 e 1 e 1 e 1 * e 2 e 2 e 2 ( e ) x y
47 Step 4: replace all occurrences of e 0 with e. After step 4: e e + e 1 e e 1 e 1 e 1 e 1 * e 2 e 2 e 2 ( e ) x y
48 Step 4: replace all occurrences of e 0 with e. After step 4: e e + e 1 e e 1 e 1 e 1 e 1 * e 2 e 2 e 2 ( e ) x y
49 Exercise 2 Consider the following ambiguous grammar for logical propositions. p 0 (Zero) 1 (One) ~ p (Negation) p + p (Disjunction) p * p (Conjunction) Now let + and * be right associative and the operators in increasing order of binding strength be : +, *, ~. Give an unambiguous grammar for logical propositions.
50 Exercise 2 Consider the following ambiguous grammar for logical propositions. p 0 (Zero) 1 (One) ~ p (Negation) p + p (Disjunction) p * p (Conjunction) Now let + and * be right associative and the operators in increasing order of binding strength be : +, *, ~. Give an unambiguous grammar for logical propositions.
51 Exercise 3 Which of the following grammars are ambiguous? b 0 b e + e e e e x s if b then s if b then s else s skip
52 Exercise 3 Which of the following grammars are ambiguous? b 0 b e + e e e e x s if b then s if b then s else s skip
53 Summary of Part 1 Syntax of a language is often specified by a contextfree grammar Derivations and parse trees are proofs that a string is accepted by a grammar. Construction of unambiguous grammars using rules of precedence and associativity.
54 Summary of Part 1 Syntax of a language is often specified by a contextfree grammar Derivations and parse trees are proofs that a string is accepted by a grammar. Construction of unambiguous grammars using rules of precedence and associativity.
55 PART 2: TOPDOWN PARSING RecursiveDescent Backtracking LeftFactoring Predictive Parsing LeftRecursion Removal First and Follow Sets Parsing tables and LL(1)
56 PART 2: TOPDOWN PARSING RecursiveDescent Backtracking LeftFactoring Predictive Parsing LeftRecursion Removal First and Follow Sets Parsing tables and LL(1)
57 Topdown parsing Topdown: begin with the start symbol and expand nonterminals, succeeding when the input string is matched. A good strategy for writing parsers: 1. Implement a syntax checker to accept or refute input strings. 2. Modify the checker to construct a parse tree straightforward.
58 Topdown parsing Topdown: begin with the start symbol and expand nonterminals, succeeding when the input string is matched. A good strategy for writing parsers: 1. Implement a syntax checker to accept or refute input strings. 2. Modify the checker to construct a parse tree straightforward.
59 RECURSIVE DESCENT A popular topdown parsing technique.
60 RECURSIVE DESCENT A popular topdown parsing technique.
61 Recursive descent A recursive descent parser consists of a set of functions, one for each nonterminal. The function for nonterminal n returns true if some prefix of the input string can be derived from n, and false otherwise.
62 Recursive descent A recursive descent parser consists of a set of functions, one for each nonterminal. The function for nonterminal n returns true if some prefix of the input string can be derived from n, and false otherwise.
63 Consuming the input We assume a global variable next points to the input string. char* next; Consume c from input if possible. int eat(char c) { if (*next == c) { next++; return 1; } return 0; }
64 Consuming the input We assume a global variable next points to the input string. char* next; Consume c from input if possible. int eat(char c) { if (*next == c) { next++; return 1; } return 0; }
65 Recursive descent Let parser(x) denote X() if X is a nonterminal eat(x) if X is a terminal For each nonterminal N, introduce: int N() { char* save = next; } for each N X 1 X 2 X n if (parser(x 1 ) && parser(x 2 ) && && parser(x n )) return 1; else next = save; return 0; Backtrack
66 Recursive descent Let parser(x) denote X() if X is a nonterminal eat(x) if X is a terminal For each nonterminal N, introduce: int N() { char* save = next; } for each N X 1 X 2 X n if (parser(x 1 ) && parser(x 2 ) && && parser(x n )) return 1; else next = save; return 0; Backtrack
67 Exercise 4 Consider the following grammar G with start symbol e. e ( e + e ) ( e * e ) v v x y Using recursive descent, write a syntax checker for grammar G.
68 Exercise 4 Consider the following grammar G with start symbol e. e ( e + e ) ( e * e ) v v x y Using recursive descent, write a syntax checker for grammar G.
69 Answer (part 1) int e() { char* save = next; if (eat('(') && e() && eat('+') && e() && eat(')')) return 1; else next = save; if (eat('(') && e() && eat('*') && e() && eat(')')) return 1; else next = save; if (v()) return 1; else next = save; return 0; }
70 Answer (part 1) int e() { char* save = next; if (eat('(') && e() && eat('+') && e() && eat(')')) return 1; else next = save; if (eat('(') && e() && eat('*') && e() && eat(')')) return 1; else next = save; if (v()) return 1; else next = save; return 0; }
71 Answer (part 2) int v() { char* save = next; if (eat('x')) return 1; else next = save; if (eat('y')) return 1; else next = save; return 0; }
72 Answer (part 2) int v() { char* save = next; if (eat('x')) return 1; else next = save; if (eat('y')) return 1; else next = save; return 0; }
73 Exercise 5 How many function calls are made by the recursive descent parser to parse the following strings? (x*x) ((x*x)*x) (((x*x)*x)*x) (See animation of backtracking.)
74 Exercise 5 How many function calls are made by the recursive descent parser to parse the following strings? (x*x) ((x*x)*x) (((x*x)*x)*x) (See animation of backtracking.)
75 Answer Number of calls is quadratic in the length of the input string. Input string Length Calls (x*x) 5 21 ((x*x)*x) 9 53 (((x*x)*x)*x) Lesson: backtracking expensive!
76 Answer Number of calls is quadratic in the length of the input string. Input string Length Calls (x*x) 5 21 ((x*x)*x) 9 53 (((x*x)*x)*x) Lesson: backtracking expensive!
77 LEFT FACTORING Reducing backtracking!
78 LEFT FACTORING Reducing backtracking!
79 Left factoring When two productions for a nonterminal share a common prefix, expensive backtracking can be avoided by leftfactoring the grammar. Idea: Introduce a new nonterminal that accepts each of the different suffixes.
80 Left factoring When two productions for a nonterminal share a common prefix, expensive backtracking can be avoided by leftfactoring the grammar. Idea: Introduce a new nonterminal that accepts each of the different suffixes.
81 Example 3 Leftfactoring grammar G by introducing nonterminal r: e ( e r v r + e ) * e ) v x y Common prefix Different suffixes
82 Example 3 Leftfactoring grammar G by introducing nonterminal r: e ( e r v r + e ) * e ) v x y Common prefix Different suffixes
83 Exercise 6 How many function calls are made by the recursive descent parser (after leftfactoring) to parse the following strings? (x*x) ((x*x)*x) (((x*x)*x)*x)
84 Exercise 6 How many function calls are made by the recursive descent parser (after leftfactoring) to parse the following strings? (x*x) ((x*x)*x) (((x*x)*x)*x)
85 Answer Number of calls is now linear in the length of input string. Input string Length Calls (x*x) 5 13 ((x*x)*x) 9 22 (((x*x)*x)*x) Lesson: leftfactoring a grammar reduces backtracking.
86 Answer Number of calls is now linear in the length of input string. Input string Length Calls (x*x) 5 13 ((x*x)*x) 9 22 (((x*x)*x)*x) Lesson: leftfactoring a grammar reduces backtracking.
87 PREDICTIVE PARSING Eliminating backtracking!
88 PREDICTIVE PARSING Eliminating backtracking!
89 Predictive parsing Idea: know which production of a nonterminal to choose based solely on the next input symbol. Advantage: very efficient since it eliminates all backtracking. Disadvantage: not all grammars can be parsed in this way. (But many useful ones can.)
90 Predictive parsing Idea: know which production of a nonterminal to choose based solely on the next input symbol. Advantage: very efficient since it eliminates all backtracking. Disadvantage: not all grammars can be parsed in this way. (But many useful ones can.)
91 Running example The following grammar H will be used as a running example to demonstrate predictive parsing. e e + e e * e ( e ) x y Example: x+y*(y+x)
92 Running example The following grammar H will be used as a running example to demonstrate predictive parsing. e e + e e * e ( e ) x y Example: x+y*(y+x)
93 Removing ambiguity Since + and * are leftassociative and * binds tighter than +, we can derive an unambiguous variant of H. e e + t t t t * f f f ( e ) x y
94 Removing ambiguity Since + and * are leftassociative and * binds tighter than +, we can derive an unambiguous variant of H. e e + t t t t * f f f ( e ) x y
95 Left recursion Problem: leftrecursive grammars cause recursive descent parsers to loop forever. int e() { char* save = next; if (e() && eat('+') && t()) return 1; next = save; if (t()) return 1; next = save; Call to self without consuming any input } return 0;
96 Left recursion Problem: leftrecursive grammars cause recursive descent parsers to loop forever. int e() { char* save = next; if (e() && eat('+') && t()) return 1; next = save; if (t()) return 1; next = save; Call to self without consuming any input } return 0;
97 Eliminating left recursion Let α denote any sequence of grammar symbols. n n α Rule 1 n' α n' n α Rule 2 n α n' where α does not begin with n Introduce new production Rule 3 n' ε
98 Eliminating left recursion Let α denote any sequence of grammar symbols. n n α Rule 1 n' α n' n α Rule 2 n α n' where α does not begin with n Introduce new production Rule 3 n' ε
99 Example 4 Running example, after eliminating leftrecursion. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
100 Example 4 Running example, after eliminating leftrecursion. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
101 first and follow sets Predictive parsers are built using the first and follow sets of each nonterminal in a grammar. The first set of a nonterminal n is the set of symbols that can begin a string derived from n. The follow set of a nonterminal n is the set of symbols that can immediately follow n in any step of a derivation.
102 first and follow sets Predictive parsers are built using the first and follow sets of each nonterminal in a grammar. The first set of a nonterminal n is the set of symbols that can begin a string derived from n. The follow set of a nonterminal n is the set of symbols that can immediately follow n in any step of a derivation.
103 Definition of first sets Let α denote any sequence of grammar symbols. If α can derive a string beginning with terminal a then a first(α). If α can derive ε then ε first(α).
104 Definition of first sets Let α denote any sequence of grammar symbols. If α can derive a string beginning with terminal a then a first(α). If α can derive ε then ε first(α).
105 Computing first sets If a is a terminal then a first(a). If there exists a production n X 1 X 2 X n and i a first(x i ) and j < i ε first(x j ) then a first(n). If n ε then ε first(n).
106 Computing first sets If a is a terminal then a first(a). If there exists a production n X 1 X 2 X n and i a first(x i ) and j < i ε first(x j ) then a first(n). If n ε then ε first(n).
107 Exercise 7 What are the first sets for each nonterminal in the following grammar. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
108 Exercise 7 What are the first sets for each nonterminal in the following grammar. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
109 Answer first( f ) = { (, x, y } first( t' ) = { *, ε } first( t ) = { (, x, y } first( e' ) = { +, ε } first( e ) = { (, x, y }
110 Answer first( f ) = { (, x, y } first( t' ) = { *, ε } first( t ) = { (, x, y } first( e' ) = { +, ε } first( e ) = { (, x, y }
111 Definition of follow sets Let α and β denote any sequence of grammar symbols. Terminal a follow(n) if the start symbol of the grammar can derive a string of grammar symbols in which a immediately follows n. The set follow(n) never contains ε.
112 Definition of follow sets Let α and β denote any sequence of grammar symbols. Terminal a follow(n) if the start symbol of the grammar can derive a string of grammar symbols in which a immediately follows n. The set follow(n) never contains ε.
113 End markers In predictive parsing, it is useful to mark the end of the input string with a $ symbol. If the start symbol can derive a string of grammar symbols in which n is the rightmost symbol then $ is in follow(n).
114 End markers In predictive parsing, it is useful to mark the end of the input string with a $ symbol. If the start symbol can derive a string of grammar symbols in which n is the rightmost symbol then $ is in follow(n).
115 Computing follow sets If s is the start symbol of the grammar then $ follow(s). If n α x β then everything in first(β) except ε is in follow(x). If n α x or n α x β and ε first(β) then everything in follow(n) is in follow(x).
116 Computing follow sets If s is the start symbol of the grammar then $ follow(s). If n α x β then everything in first(β) except ε is in follow(x). If n α x or n α x β and ε first(β) then everything in follow(n) is in follow(x).
117 Exercise 8 What are the follow sets for each nonterminal in the following grammar. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
118 Exercise 8 What are the follow sets for each nonterminal in the following grammar. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
119 Answer follow( e' ) = { $, ) } follow( e ) = { $, ) } follow( t' ) = { +, $, ) } follow( t ) = { +, $, ) } follow( f ) = { *, +, ), $ }
120 Answer follow( e' ) = { $, ) } follow( e ) = { $, ) } follow( t' ) = { +, $, ) } follow( t ) = { +, $, ) } follow( f ) = { *, +, ), $ }
121 Predictive parsing table For each nonterminal n, a parse table T defines which production of n should be chosen, based on the next input symbol. for each production n α for each a first(α) add n α to T[n, a] if ε first(α) then for each b follow(n) add n α to T[n, a]
122 Predictive parsing table For each nonterminal n, a parse table T defines which production of n should be chosen, based on the next input symbol. for each production n α for each a first(α) add n α to T[n, a] if ε first(α) then for each b follow(n) add n α to T[n, a]
123 Exercise 9 Construct a predictive parsing table for the following grammar. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
124 Exercise 9 Construct a predictive parsing table for the following grammar. e t e' e' + t e' ε t f t' t' * f t' ε f ( e ) x y
125 LL(1) grammars If each cell in the parse table contains at most one entry then the a nonbacktracking parser can be constructed and the grammar is said to be LL(1). First L: lefttoright scanning of the input. Second L: a leftmost derivation is constructed. The (1): using one input symbol of lookahead to decide which grammar production to choose.
126 LL(1) grammars If each cell in the parse table contains at most one entry then the a nonbacktracking parser can be constructed and the grammar is said to be LL(1). First L: lefttoright scanning of the input. Second L: a leftmost derivation is constructed. The (1): using one input symbol of lookahead to decide which grammar production to choose.
127 Exercise 10 Write a syntax checker for the grammar of Exercise 9, utilising the predictive parsing table. int e() {... } It should return a nonzero value if some prefix of the string pointed to by next conforms to the grammar, otherwise it should return zero.
128 Exercise 10 Write a syntax checker for the grammar of Exercise 9, utilising the predictive parsing table. int e() {... } It should return a nonzero value if some prefix of the string pointed to by next conforms to the grammar, otherwise it should return zero.
129 Answer (part 1) int e() { if (*next == 'x') return t() && e1(); if (*next == 'y') return t() && e1(); if (*next == '(') return t() && e1(); return 0; } int e1() { if (*next == '+') return eat('+') && t() && e1(); if (*next == ')') return 1; if (*next == '\0') return 1; return 0; }
130 Answer (part 1) int e() { if (*next == 'x') return t() && e1(); if (*next == 'y') return t() && e1(); if (*next == '(') return t() && e1(); return 0; } int e1() { if (*next == '+') return eat('+') && t() && e1(); if (*next == ')') return 1; if (*next == '\0') return 1; return 0; }
131 Answer (part 2) int t() { if (*next == 'x') return f() && t1(); if (*next == 'y') return f() && t1(); if (*next == '(') return f() && t1(); return 0; } int t1() { if (*next == '+') return 1; if (*next == '* ) return eat('*') && f() && t1(); if (*next == ')') return 1; if (*next == '\0') return 1; return 0; }
132 Answer (part 2) int t() { if (*next == 'x') return f() && t1(); if (*next == 'y') return f() && t1(); if (*next == '(') return f() && t1(); return 0; } int t1() { if (*next == '+') return 1; if (*next == '* ) return eat('*') && f() && t1(); if (*next == ')') return 1; if (*next == '\0') return 1; return 0; }
133 Answer (part 3) int f() { if (*next == 'x') return eat('x'); if (*next == 'y') return eat('y'); if (*next == '(') return eat('(') && e() && eat(')'); return 0; } (Notice how backtracking is not required.)
134 Answer (part 3) int f() { if (*next == 'x') return eat('x'); if (*next == 'y') return eat('y'); if (*next == '(') return eat('(') && e() && eat(')'); return 0; } (Notice how backtracking is not required.)
135 Predictive parsing algorithm Let s be a stack, initially containing the start symbol of the grammar, and let next point to the input string. while (top(s)!= $) if (top(s) is a terminal) { if (top(s) == *next) { pop(s); next++; } else error(); } else if (T[top(s), *next] == X Y 1 Y n ) { pop(s); push(s, Y n Y 1 ) /* Y 1 on top */ }
136 Predictive parsing algorithm Let s be a stack, initially containing the start symbol of the grammar, and let next point to the input string. while (top(s)!= $) if (top(s) is a terminal) { if (top(s) == *next) { pop(s); next++; } else error(); } else if (T[top(s), *next] == X Y 1 Y n ) { pop(s); push(s, Y n Y 1 ) /* Y 1 on top */ }
137 Exercise 11 Give the steps that a predictive parser takes to parse the following input. x + x * y For each step (loop iteration), show the input stream, the stack, and the parser action.
138 Exercise 11 Give the steps that a predictive parser takes to parse the following input. x + x * y For each step (loop iteration), show the input stream, the stack, and the parser action.
139 Acknowledgements Plus Stanford University lecture notes by Maggie Johnson and Julie Zelenski.
140 Acknowledgements Plus Stanford University lecture notes by Maggie Johnson and Julie Zelenski.
141 APPENDIX
142 APPENDIX
143 Chomsky hierarchy Let t range over terminals, x and z over nonterminals and, β and γ over sequences of terminals, nonterminals, and ε. Grammar Unrestricted Valid productions α β ContextSensitive α x γ α β γ ContextFree Regular x β x t x t z x ε
144 Chomsky hierarchy Let t range over terminals, x and z over nonterminals and, β and γ over sequences of terminals, nonterminals, and ε. Grammar Unrestricted Valid productions α β ContextSensitive α x γ α β γ ContextFree Regular x β x t x t z x ε
145 BackusNaur Form BNF is a standard ASCII notation for specification of contextfree grammars whose terminals are ASCII characters. For example: <exp> ::= <exp> "+" <exp> <exp> "" <exp> <var> <var> ::= "x" "y" The BNF notation can itself be specified in BNF.
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