[email protected] chapter 3 context-free grammars and parsing ambiguous grammar gang s....
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Chapter 3
Context-Free Grammars and Parsing
Ambiguous Grammar
Gang S. LiuCollege of Computer Science &
TechnologyHarbin Engineering University
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Ambiguous Grammar
• The string 34 – 3 * 42 has two different parse trees and syntax trees.
exp → exp op exp | (exp) | number
op → + | - | *
*
- 42
34 3
-
34 *
3 42
A grammar that generates a string with two distinct parse trees is called an ambiguous grammar.
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Two Ways to Deal with Ambiguity
1. Disambiguating rule• State a rule that specifies in each ambiguous case
which of the trees is correct.
• It corrects the ambiguity without changing and complicating the grammar.
• Syntactic structure of the language is not given by the grammar alone.
2. Change grammar into a form that forces the construction of a correct parse tree.
• In both cases we must decide which of the trees are correct.
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Ambiguous Grammar
• The string 34 – 3 * 42 has two different parse trees and syntax trees.
exp → exp op exp | (exp) | number
op → + | - | *
*
- 42
34 3
-
34 *
3 42
A grammar that generates a string with two distinct parse trees is called an ambiguous grammar.
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Precedence
• Some operations have precedence over other operations.– Multiplication has precedence over addition.
• To handle the precedence of operations in the grammar, we group the operators into groups of equal precedence.
• We must write a different rule for each precedence.
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Exampleexp → exp op exp | (exp) | numberop → + | - | *
exp → exp addop exp | termaddop → + | -term → term mulop term | factormulop → *factor → (exp) | number
*
- 42
34 3
-
34 *
3 42 √√××
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Ambiguous
(34 – 3) – 42 = –11 34 – (3 – 42) = 73
34 – 3 – 42
(34 – 3) – 42
√√ ××
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Associativity• Some operation (like subtraction) are left
associative.– A series of subtraction operations is performed from
left to right.
• exp → exp addop exp | term– Recursion on both sides of the operator allows either
side to match repetitions of the operator in a derivation.
• exp → exp addop term | term– The right recursion is replaced with the base case,
forcing the repetitive matches on the left side
– Left recursion makes addition and subtraction left associative.
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Exampleexp → exp addop term | termaddop → + | -term → term mulop factor | factormulop → *factor → (exp) | number
34- 3* 42 has a unique parse tree
exp
exp addop term
term - term mulop factor
factor factor * number
number number
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Dangling else Problem
• if (0) if (1) other else other has two parse trees with two meaning
if (0) if (1) other else other and
if (0) if (1) other else other
statement → if-stmt | other
if-stmt → if (exp) statement
| if (exp) statement else statement
exp → 0 | 1
√√××
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Dangling else Problem
• if (0) if (1) other else other has two parse trees with two meaning
if (0) if (1) other else other and
if (0) if (1) other else other
1. Modify the grammar.
2. Disambiguating rule: the most closely nested rule.
statement → if-stmt | other
if-stmt → if (exp) statement
| if (exp) statement else statement
exp → 0 | 1
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Dangling else Problem
statement → matched-stmt | unmatched-stmt
matched-stmt → if (exp) matched-stmt else
matched-stmt | other
unmatched-stmt → if (exp) statement
| if (exp) matched-stmt else unmatched-stmt
exp → 0 | 1
if (0) if (1) other else other
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Inessential Ambiguity
• Sometimes a grammar may be ambiguous and yet always produce unique abstract syntax trees.
• Example:
( a + b ) + c = a + ( b + c )
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Extended BNF Notation (EBNF)
• { } repetitions
A → β {α} and A → {α} β
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Left and Right Recursion
• Left recursive grammar:
• A → A α | β – Equivalent to β α *
• A → β {α}
• Right recursive grammar:
• A → α A | β - Equivalent to α * β
• A → {α} β
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Extended BNF Notation (EBNF)
• { } repetitions
A → β {α} and A → {α} β• [ ] optional constructs
statement → if-stmt | other
if-stmt → if (exp) statement [ else statement ]
exp → 0 | 1
statement → if-stmt | otherif-stmt → if (exp) statement | if (exp) statement else statementexp → 0 | 1
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Syntax Diagrams
• Graphical representations for visually representing EBNF rules are called syntax diagrams.
• They consist of boxes representing terminals and nonterminals, arrowed lines representing sequencing and choices, and nonterminal labels for each diagram representing the grammar rule defining that nonterminal.
• A round or oval box is used to indicate terminals in a diagram, while a square or rectangular box is used to indicate nonterminals.
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Syntax Diagrams(cont)
• As an example, consider the grammar rule• factor → ( exp ) | number• This is written as a syntax diagram in the
following way:
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Syntax Diagrams(cont)
• Syntax diagrams are written from the EBNF rather than the BNF, so we need diagrams representing repetition and optional constructs. Given a repetition such as
• A → { B } • The corresponding syntax diagram is usually
drawn as follow:
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Syntax Diagrams(cont)
• An optional construct such as• A → [ B ] • Is drawn as:
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Example
exp → exp addop term | termaddop → + | –term → term mulop factor | factormulop → *factor → (exp) | number
exp → term { addop term }addop → + | –term → factor { mulop factor }mulop → *factor → (exp) | number
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Example
statement → if-stmt | other
if-stmt → if (exp) statement [ else statement ]
exp → 0 | 1
statement → if-stmt | otherif-stmt → if (exp) statement | if (exp) statement else statementexp → 0 | 1
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Context-free Grammar for TINYprogram → stmt-sequencestmt-sequence → stmt-sequence ; statement | statementstatement → if-stmt | repeat-stmt | assign-stmt| read-stmt | write-stmtif-stmt → if exp then stmt-sequence end | if exp then stmt-sequence else stmt-sequence endrepeat-stmt → repeat stmt-sequence until expassign-stmt → identifier := expread-stmt → read identifierwrite-stmt → write expexp → simple-exp comp-op simple-exp | simple-expcomp-op → < | =simple-exp → simple-exp addop term | termaddop → + | -term → term mulop factor | factormulop → * | /factor → (exp) | number | identifier
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Factorial Program
{ Sample program in TINY language - computes factorial }read x; { input an integer }if 0 < x then { don't compute if x <= 0 } fact := 1; repeat fact := fact * x; x := x - 1 until x = 0; write fact { output factorial of x }end
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Homework
1.2 1.3 1.7
2.1 2.2 2.4 2.5 2.8 2.12 2.17 2.24
3.3 3.5 3.6 3.24
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Homework
3.5 Write a grammar for Boolean expressions that includes the constants true and false, the operators and, or, and not, and parentheses. Be sure to give or a lower precedence than and and and a lower precedence than not and to allow repeated not’s, as in the Boolean expression not not true. Also be sure your grammar is not ambiguous.
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Homework
3.6 Consider the following grammar representing simplified LISP-like expressions:
a. Write a leftmost and a rightmost derivation for the string ( a 23 (m x y) ) .
b. Draw a parse tree for the string of part(a).
lexp → atom | listatom → number | identifierlist → (lexp-seq) lexp-seq → lexp-seq lexp | lexp
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Homework
3.24 For the TINY program
a. Draw the TINY parse tree.
b. Draw the TINY syntax tree.
read x;x := x + 1;write x