Question

我在这个Bison计划中遇到了麻烦。它必须通过将它们乘以2 ^ n来接收一个1s和0s的字符串，其周期类似于“101.101”。例如：

"101.101" = (1*2^2)+(0*2^1)+(1*2^0)+(1*2^-1)+(0*2^-2)+(1*2^-3)=5.625

这一点说明战俘何时是积极的还是消极的。我有以下语义行为：

S→ L.R 
 S→ L  
 L → L1 B   
 L → B  
 R → R1 B  
 R → B   
 B→ 0  
 B → 1  
 Sematic Rules  
 L.pos=0;R.pos=-1;S.val=L.val+R.val  
 L.pos=0;S.val=L.val;  
 L1.pos = L.pos + 1; B.pos = L.pos; L.val = L1.val + B.val;  
 B.pos = L.pos; L.val = B.val;  
 R1.pos = R.pos - 1; B.pos = R.pos; L.val = L1.val + B.val;
 B.pos = R.pos; L.val = B.val;  
 B.val=0;  
 B.val = 1*2^B.pos;

我现在遇到的问题是我不知道为什么.pos变量不起作用，它们的值总是为0.我的野牛代码是：

%{
#include <string.h>
#include <stdio.h>
#include<stdlib.h>
void yyerror (char *string);

%}
%union {
    struct named_for_discussion_below {
        int pos;
        int val;
    } pair;
}

%token DOT
%token ZERO
%token ONE
%token l1
%token r1
%type <pair> b l r s;


%%
x: s {/*printf(" the number is %d \n",$1);*/}
 ;

s: l DOT r {$1.pos=0;$3.pos=-1;$$.val=$1.val+$3.val;/*printf(" the both numbers are %d and       %d\n",$1,$3);*/}
| l {$1.pos=0;$$.val=$1.val;/*printf(" the numbers is %d \n",$1);*/}
 ;

 l: l b {$1.pos = $$.pos + 1; $2.pos = $$.pos; $$.val = $1.val + $2.val;printf(" the   number is left,  l pos is %d and l val is %d \n", $$.pos, $$.val);}
| b   {$1.pos = $$.pos; $$.val = $1.val;printf(" the number is left,  l pos is %d and l  val is %d \n", $$.pos, $$.val);}
 ;

r: r b {$1.pos = $$.pos - 1; $2.pos = $$.pos; $$.val = $1.val + $2.val;printf(" the number     is right, r pos is %d and r val is %d \n", $$.pos, $$.val);}
| b   {$1.pos = $$.pos; $$.val = $1.val; printf(" the number is right, r pos is %d and r val is %d \n", $$.pos, $$.val);}

 ;

b: ZERO {$$.val = 0; printf(" the number is 0, val is %d and pos is %d \n",$$.val,$$.pos);}
| ONE {$$.val = 1*2^($$.pos); printf(" the number is 1, val is %d and pos is %d  \n",$$.val,$$.pos);}
 ;

%%
#include "lex.yy.c"

void yyerror (char *string){
  printf ("%s",string);
}

int main (){

    yyparse();
}

和lex文件是：

%{
#include <stdio.h>
#include <math.h>
#include "y.tab.h"
%}
BINARY [0-1]
%%
"1" {return ONE;}
"0" {return ZERO;}
"." {return DOT;}
%%

Answer 1

yacc中的属性始终是合成属性，其值从叶子向上传播到解析树的根，而不是向下传播。

如果要使用继承的属性，则需要使用btyacc之类的工具（您可以获得更新版本here）。这允许您编写代码，如：

%{
#include <string.h>
#include <stdio.h>
#include<stdlib.h>
%}
%union {
    double  val;
    int     pos;
}

%token DOT
%token ZERO
%token ONE
%token l1
%token r1
%type <val> b(<pos>) l(<pos>) r(<pos>) s;


%%
x: s {printf(" the number is %f \n",$1);}
 ;

s: l(0) DOT r(-1) {$$=$1+$3; /*printf(" the both numbers are %f and %f\n",$1,$3);*/}
 | l(0) {$$=$1; /*printf(" the numbers is %f \n",$1);*/}
 ;

l($pos): l($pos+1) b($pos) { $$ = $1 + $2; printf(" the   number is left,  l pos is %d and l val is %f \n", $pos, $$);}
       | b($pos)   { $$ = $1; printf(" the number is left,  l pos is %d and l  val is %f \n", $pos, $$);}
       ;

r($pos): b($pos) r($pos-1) { $$ = $1 + $2; printf(" the number     is right, r pos is %d and r val is %f \n", $pos, $$);}
       | b($pos)   { $$ = $1; printf(" the number is right, r pos is %d and r val is %f \n", $pos, $$);}
       ;

b($pos): ZERO { $$ = 0; printf(" the number is 0, val is %f and pos is %d \n",$$,$pos);}
       | ONE { $$ = pow(2.0, $pos); printf(" the number is 1, val is %f and pos is %d  \n",$$,$pos);}
       ;

%%
#include "lex.yy.c"

void yyerror (const char *string, ...){
  printf ("%s",string);
}

int main (){
    yyparse();
}

请注意，我还将val更改为double，因为int只能保留整数。我还将其更改为使用pow进行取幂（C中的^是xor）。

Answer 2

在您的操作中，您希望从$ 1和$ 2分配给$$。因此，$$应位于'='的左侧，$ 1和右侧的$ 2。所以...

l: l b {$1.pos = $$.pos + 1; $2.pos = $$.pos; $$.val = $1.val + $2.val;printf(" the   number is left,  l pos is %d and l val is %d \n", $$.pos, $$.val);}
             ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
    | b   {$1.pos = $$.pos; $$.val = $1.val;printf(" the number is left,  l pos is %d and l  val is %d \n", $$.pos, $$.val);}
           ^^^^^^^^^^^^^^^^ 
     ;

还有其他几种相同的品种。

r: r b {$1.pos = $$.pos - 1; $2.pos = $$.pos; $$.val = $1.val + $2.val;printf(" the number     is right, r pos is %d and r val is %d \n", $$.pos, $$.val);}
        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
| b   {$1.pos = $$.pos; $$.val = $1.val; printf(" the number is right, r pos is %d and r val is %d \n", $$.pos, $$.val);}
        ^^^^^^^^^^^^^^^

我不清楚你的表达式应该是什么，但$$应该是1美元和2美元的函数。

Answer 3

$$是语义操作的结果，因此将其作为赋值的来源是没有意义的。此外，您不能向下传递值，您必须使用其他方法将值传递给更基本的规则（如全局变量）。

野牛语义规则变量

3 个答案: