Job 11 LL (1) grammar is determined, recursive descent parser

1. grammar  G (S):

1)S -> AB

2)A ->Da|ε

3)B -> cC

4)C -> aADC |ε

5)D -> b|ε

Verify grammar  G (S) is not  LL (1) grammar?

FIRST集:

FOLLOW集:

FIRST(A) = { b,a, ε }

FIRST(C) = { a }

FIRST(D) = { b }

FIRST(Da) = { b,a }

FIRST (e) = {e}

FIRST(aADC) = {a}

FIRST(b) = { b }

FOLLOW(A) = { c,b,a,# }

FOLLOW(C) = { # }

FOLLOW(D) = { a,# }

 

SELECT set:

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

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

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

SELECT(C -> ε) = FIRST(ε) - {ε} U FOLLOW(aADC) = FOLLOW(C) = { # }

SELECT(D ->b) = FIRST(b) = { b }

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

 

即有:  SELECT(A -> Da)∩SELECT(A -> ε)≠∅

SELECT(C ->aADC)∩SELECT(C -> ε)=∅

SELECT(D ->b)∩SELECT(D -> ε)=∅

 

That grammar G (S) is not LL (1) syntax.

 

2. (last job) after the elimination of left recursion whether the expression grammar is LL (1) grammar?

 E -> E+T | T

 T -> T*F | F

 F -> (E) | i

 

Eliminate left recursion:

E -> TE '

E '-> + TE' |

T -> FT’

T’ -> *FT’ | ε 

F -> (E) | i

 

FIRST集:

FOLLOW集:

FIRST (E ') = {+, e}

FIRST(T’) = { * , ε }

FIRST(F) = { ( , i }

FIRST(+TE') = {+}

FIRST (e) = {e}

FIRST(*FT') = {*}

FIRST((E)) = { ( }

FIRST(i) = {i}

FOLLOW(E’) = { ) , # }

FOLLOW(T’) = { + , ) ,# }

FOLLOW(F) = { * , + , ) ,# }

SELECT set:

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

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

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 }

 

 

Namely:  the SELECT (E '-> + the TE') ∩SELECT (E '-> [epsilon]) = ∅

SELECT(T’ -> *FT’)∩SELECT(T’ -> ε)=∅

SELECT(F -> (E))∩SELECT(F -> i)=∅

 

That is, the grammar is LL (1) syntax.

 

3. The connection 2, if it is LL (1) grammar, write its recursive descent parser code.

void ParseE(){

  if(lookhead==(,i){

    Frset ();

    ParseE’();

    break;

  }else{

    print(“syntax error \n”);

    exit(0);

  }

}

 

void ParseE’(){

  Switch(lookhead){

    case +:

      Match Token (+);

      Frset ();

      ParseE’();

      break;

    case ),#:

      break;

    default:

      print(“syntax error \n”);

      exit(0);

    }

}

 

void ParseT(){

  if(lookhead==(,i){

    ParseF();

    Frset '();

    break;

  }else{

    print(“syntax error \n”);

    exit(0);

  }

}

 

void ParseT’(){

  Switch(lookhead){

    case *:

      MatchToken(*);

      ParseF();

      ParseT’();

      break;

    case +,),#:

      break;

     default:

       print(“syntax error \n”);

       exit(0);

    }

}

 

void ParseF’(){

  Switch(lookhead){

    case (:

      MatchToken(();

      ParseE();

      MatchToken());

      break;

    case i:

      MatchToken(i);

      break;

    default:

      print(“syntax error \n”);

      exit(0);

    }

}

 

 4.加上实验一的词法分析程序,形成可运行的语法分析程序,分析任意输入的符号串是不是合法的表达式。

 

Guess you like

Origin www.cnblogs.com/seele233/p/11896700.html