11.20LL(1)文法的判断,递归下降分析程序

1. 文法 G(S):

(1)S -> AB

(2)A ->Da|ε

(3)B -> cC

(4)C -> aADC |ε

(5)D -> b|ε

验证文法 G(S)是不是 LL(1)文法?

 答:

FIRST集:

       FIRST(A) = { b , a , ε}

       FIRST(C) = { a , ε}

       FIRST(D) = { b , ε}

FOLLOW集:

  FOLLOW (A) = { c , b , a , ε}

       FOLLOW (C) = { # }

       FOLLOW (D) = { a , #}

SELECT集:

       SELECT( A -> Da ) = FIRST( Da ) = { b , a }

       SELECT( A -> Da) = FIRST(Da) = { b, a }

  SELECT( A -> ε) = FOLLOW( A) = { c, b, a, # }

  SELECT( C -> aADC) = FIRST( aADC) = { a }

  SELECT( C -> ε) = FOLLOW(C) = { # }

  SELECT( D -> b) = FIRST(b) = { b }

  SELECT( D -> ε ) =FOLLOW(D) = { a, # }

因为SELECT( A -> Da) ∩ SELECT( A -> ε) = { a } ≠ ∅,所以文法G(S)不是 LL(1)文法。

2.(上次作业)消除左递归之后的表达式文法是否是LL(1)文法?

1.将以下文法消除左递归,分析符号串 i*i+i 。

   并分别求FIRST集、FOLLOW集,和SELECT集

     E -> E+T | T

     T -> T*F | F

     F -> (E) | i

 答:

消除左递归:

E -> TE'

      E' -> +TE' | ε 

      T -> FT'

      T' -> *FT' | ε 

      F -> (E) | i

FIRST集:

  FIRST(E) = { ( , i }

  FIRST(E') = {+ , ε }

  FIRST(T) = { ( , i }

  FIRST(T') = { * , ε }

  FIRST(F) = { ( , i }

FOLLOW集:

       FOLLOW(E) = { ) , # }

       FOLLOW(E') = { ) , # }

       FOLLOW(T) = { + , ) ,#}

       FOLLOW(T') = {+ , ) ,#}

       FOLLOW(F) = {* , + , ) ,#}

SELECT集:

  SELECT (E -> TE') = FIRST(TE') = { ( , i }

       SELECT(E' -> +TE') = FIRST(+TE') = { + }

       SELECT(E' -> ε) = FIRST(ε) - {ε} U FOLLOW(E') = FOLLOW(E') = { ) , # }

       SELECT(T -> FT') = FIRST(FT') = { ( , i }

       SELECT(T' -> *FT') = FIRST(*FT') = { * }

       SELECT(T' -> ε) = FIRST(ε) - {ε} U FOLLOW(T') = FOLLOW(T') = { + , ) ,# }

       SELECT(F -> (E)) = FIRST((E)) = { ( }

       SELECT(F -> i) = FIRST(i) = { i }

因为SELECT(E' -> +TE') ∩ SELECT(E' -> ε) = ∅,

  SELECT(T' -> *FT') ∩ SELECT(T' -> ε) = ∅,

  SELECT(F -> (E)) ∩ SELECT(F -> (E))   = ∅,

所以该文法是 LL(1)文法

3.2,如果是LL(1)文法,写出它的递归下降语法分析程序代码。

E()

    T();

       E'();

     }

E'()

T()

T'()

F()

 答:

void ParseE(){

      if(lookhead =='(' || lookhead == 'i' ){

     ParseT();

     ParseE'();

}

      else{

        printf("syntx error ");

    exit(0);

  }

}

void ParseE'(){

  switch(lookahead){

           case '+':

                  MatchToken( '+' );

                  ParseT();

                  ParseE'();

                  break;

           case ')' , '#':

                  break;

           default:

                  printf("syntx error ");

                  exit(0);

      }

}

void ParseT(){

  if(lookhead == '(' || lookhead == 'i' ){

    ParseF();

    ParseT'();

  }

  else{

    printf("syntx error ");

    exit(0);

  }

}

void ParseT'(){

  switch(lookahead){
           case '*':

                  MatchToken( '*' );

                  ParseF();

                  ParseT'();

                  break;

           case '+' , ')' , '#':

                  break;

           default:

                  printf("syntx error ");

                  exit(0);

    }

}

void ParseF(){

  switch(lookahead){

           case '(':

                  MatchToken( '(' );

                  ParseE();

                  MatchToken( ')' );

                  break;

           case 'i':

                  MatchToken( 'i' );

                  break;

           default:

                 printf("syntx error ");

                  exit(0);

    }

}

原文地址:https://www.cnblogs.com/Azan1999/p/11896882.html