Syntactical analysis Syntactical analysis Context-free grammars A - - PowerPoint PPT Presentation

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Syntactical analysis Syntactical analysis Context-free grammars A - - PowerPoint PPT Presentation

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Syntactical analysis Syntactical analysis Context-free grammars A context-free grammar is a 4-tuple G = ( N , , P , S ) g ( ) Derivations 1. N is a set of non terminals Parse Trees 2. is a set of terminals (disjoint from N )

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SLIDE 1

Syntactical analysis

  • Context-free grammars
  • Derivations
  • Parse Trees
  • Left-recursive grammars
  • Top-down parsing
  • non-recursive predictive parsers
  • construction of parse tables
  • Bottom-up parsing
  • shift/reduce parsers
  • LR parsers
  • GLR parsers
  • SGLR parsers

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Syntactical analysis

A context-free grammar is a 4-tuple G = (N, Σ, P, S) g ( )

  • 1. N is a set of non terminals
  • 2. Σ is a set of terminals (disjoint from N)
  • 3. P is a subset of (N  Σ)*  N

An element (α, A)  P is called a production A A A ::= α or α  A

  • 4. S  N is the start symbol

The sets N, Σ, P are finite

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Syntactical analysis

A context-free grammar can be consider as a simple rewrite system: A   if  A  P (, ,   (NΣ)*, A  N) Example N = {E}, Σ = {+,*,(,),-, a}, S = E, P = { E + E  E E * E  E E E  E ( E )  E

  • E  E

} a  E} Derivation: E  -E  -(E) -(E+E) -(a+E) -(a+a)

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Syntactical analysis

The language L(G) generated by the context-free grammar G = (N, Σ, P, S) is: L(G) = {w  Σ* | S + w} A sentence w  L(G) contains only terminals A sentence w  L(G) contains only terminals A sentential form  is a string of terminals and non- t i l hi h b d i d f S terminals which can be derived from S: S *  with   (N  Σ)* A sentence in L(G) is a sentential form in which no A sentence in L(G) is a sentential form in which no non-terminals occur

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SLIDE 2

Syntactical analysis

Left/right derivations Th h i t b d f h d i ti t

  • There are choices to be made for each derivation step:
  • which non-terminal must be replaced?
  • which alternative of the selected non-terminal must be applied?
  • Always selecting the leftmost non-terminal in the sentential form gives
  • Always selecting the leftmost non-terminal in the sentential form gives

a leftmost derivation: lm

  • There exists also a rightmost derivation: rm
  • Consider the context-free grammar for expressions:
  • Leftmost derivation for -(a+a)

E  -E  -(E) -(E+E) -(a+E) -(a+a) Ri ht t d i ti f

  • Rightmost derivation for -(a+a)

E  -E  -(E) -(E+E) -(E+a) -(a+a)

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Syntactical analysis

A parse tree for a context-free grammar is a G = (N, Σ, P, S) tree: 1 The root is labeled ith S (the start non terminal)

  • 1. The root is labeled with S (the start non-terminal)
  • 2. Each leaf is labeled with a terminal ( Σ)or ε
  • 3. All other nodes are labeled with a non-terminal

If A is the label of a node and X1,…,Xn are the labels of the children (from left to right) then X1,…,Xn  A

1 n

must be a production rule in G (with Xi is either a terminal or a non-terminal) Special case: ε  A with label A which has exactly one child with Special case: ε  A with label A which has exactly one child with label ε

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Syntactical analysis

Example:

E  -E  -(E) -(E+E) -(a+E) -(a+a) E  E  E  E  E  E  E

  • E -

E - E - E - E - E ( E ) ( E ) ( E ) ( E ) ( E ) E + E E + E E + E a a a

The parse tree abstracts from the derivation order

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p

Syntactical analysis

Acceptor and parser For each grammar G there exists a decision procedure (acceptor) AG for L(G): AG: STRING  {true, false} such that AG(w) = true  w  L(G) AG(w) = true  w  L(G) A parser is an acceptor which constructs a parse tree as well. f

  • A top-down parser constructs the tree starting from the root
  • A bottom-up parser constructs the tree starting from the leafs

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SLIDE 3

Syntactical analysis

During parsing the following problems may occur:

  • The grammar is ambiguous
  • The grammar is left recursive
  • The grammar contains cycles
  • The grammar contains cycles

A grammar G is ambiguous if one word w  L(G) has at g g ( ) least two parse trees

  • Expression grammar without associativities and

priorities priorities

  • Dangling else problem

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Syntactical analysis

  • A grammar is immediate left recursive if the

g grammar contains a rule of the form   A

  • A grammar is left recursive if there exists a non-

terminal A and a string   (N  Σ)* such that A  A terminal A and a string   (N  Σ)* such that A  A

  • This means that after one or more steps in a

derivation an occurrence of A reduces again to an g

  • ccurrence of A without recognizing any terminal in

the input sentence.

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Syntactical analysis

Examples of indirect left recursion B   A A   B

  • r worse

B   A D A   B D A   B ε  D  G  D It is easy to remove left recursion from a context-free grammar

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Syntactical analysis

Elimination of left recursion A   A   A produce the sentential forms:  n A t f i l t ( l ft i ) l A set of equivalent (non left recursive) rules are  A’  A  A’  A’  A  A ε  A’

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SLIDE 4

Syntactical analysis

Example: (1) E + T  E (immediate left rec.) (2) T  E (3) T * F  T (immediate left rec.) ( ) ( ) (4) F  T (5) ( E )  F (6) a  F (6) a  F Applying the left recursion elimination transformation: (1) E   E (with  = + T) (2)   E (with  = T)

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Syntactical analysis

Example: (1’) T E’  E (2’) + T E’  E’ (2’’) ε  E’ th f the same for: (3’) F T’  T (4’) * F T’  T’ (4 ) F T  T (4’’) ε  T’

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Syntactical analysis

Indirect left recursion elimination

  • Suppose we have a rule of the form

B   A 1  B 2  B … n  B

  • The rule B   A is now transformed into:

1   A 2   A

2

… n   A

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Syntactical analysis

This process is repeated until either

  • t   A; the process stops, or
  • A  A; the immediately left recursion elimination

l b li d rule can be applied

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SLIDE 5

Syntactical analysis

Left factorization In general it is efficient to mo e the difference bet een the

  • In general it is efficient to move the difference between the

alternatives of a non-terminal as far as possible to the left

  • Productions of the form

 1  A  1  A  2  A …  n  A

n

  • Are equivalent with

 A’  A 1  A’ … n  A’

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Syntactical analysis

Example if b then S else S  S if b then S  S Only at the occurrence of else it can be decided which Only at the occurrence of else it can be decided which alternative should have been selected An equivalent grammar is if b then S S’  S else S  S’ else S  S ε  S’

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Syntactical analysis

Top-down parsing

  • A top-down parser “guesses” the next alternative to be

recognized, and verifies whether this alternative can be recognized in the input. If not, another alternative will be tried

  • Constructs the parse tree starting at the root
  • Constructs the parse tree starting at the root
  • Finds the leftmost derivation of the sentence

Alternative types of top down parsers:

  • Alternative types of top-down parsers:
  • recursive descent parser with backtracking
  • recursive descent parser without backtracking (“predictive

parser”) parser )

  • non-recursive predictive parser (uses push-down automaton)

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Syntactical analysis

Recursive descent parser with backtracking

  • Grammar

c A d  S a  A a b  A

  • Parser
  • Parser

bool proc S() { if input = ‘c’ then inptr +:= 1; if A() then if input = ‘d’ then inptr +:= 1; check for EOF;return(true) fi fi fi; fi; return(false) }

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SLIDE 6

Syntactical analysis

bool proc A() { isave := inptr; isave := inptr; if input = ‘a’ then inptr +:= 1; if input = ‘b’ then inptr +:= 1; return(true) fi fi; inptr := isave; inptr := isave; if input = ‘a’ then inptr +:= 1; return(true) else return(false) fi }

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Syntactical analysis

Recursive descent without backtracking

  • For ``some'' context-free grammars a recursive descent

grammar without backtracking can be derived Grammar: T E’  E + T E’  E’ ε  E’ ε  E F T’  T * F T’  T’ ε  T’ ( E )  F a  F

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Syntactical analysis

Recursive descent without backtracking

  • Parser:

proc E() { proc E’() { T(); if input = ‘+’ E’() then inptr +:= 1; T(); E’() } fi } proc T() { proc T’() { F(); if input = ‘*’ T’() then inptr +:= 1; F(); T’() } fi }

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Syntactical analysis

Recursive descent without backtracking proc F() { if input = ‘a’ then inptr +:= 1 else if input = ‘(’ else if input ( then inptr +:= 1; E(); if input = ‘)’ then inptr +:= 1 then inptr +: 1 else ERROR() fi else ERROR() fi fi fi }

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