c - yacc在reduction中丢失值

Question

我正在研究此语法以构建用于类型检查或类似功能的 SDD。我昨天花了很多时间研究数据结构和解析操作，但我总是遇到段错误。在我看来，YACC(bison) 在减少过程中正在失去值(value)。

因此我决定用更简单的操作构建一个更简单的语法。似乎值(value)在一次又一次的减少中丢失了，或者我做错了什么？词法分析器部分与本示例无关，因此我将其省略。

遵循语法及其操作以及结果与预期结果..

D:   T VAR SEMICOLON D              {
                                    printf("processing D -> T var ; D\n");
                                    printf("\tvalue of T is %f\n", $1);
                                }
|/*empty*/                      {
                                    printf("processing D -> empty\n");
                                }
;

T:  B                               {
                                    printf("processing B inside T\n");
                                    printf("\tvalue of B is %f\n", $1);
                                } 

C                               {   printf("processing C inside T\n");
                                    printf("processing T-> B C\n");
                                    printf("\tvalue of B is %f\n", $1);
                                    printf("\tvalue of C is %f\n", $<dbl>2);
                                    $$ = $1 + $<dbl>2;

                                }
| RECORD '{' D '}'              {   printf("processing record { D }\n");}
;

B:   INT                            {   printf("processing B -> int\n");
                                    $$ = 1;
                                }
| FLOAT                         {   printf("processing B -> float\n");
                                    $$ = 1;
                                }
;

C:  /*empty*/                       {   printf("processing C -> empty\n");
                                    printf("\tsetting C to be equal to 1\n");
                                    $$=1;
                                }
| LBRACK NUM RBRACK C           {   int n = $2;
                                    printf("processing C -> [%d] C\n", n);
                                    double d = $4;
                                    printf("\tprevious C value is %f\n", d);
                                    double f = d+ 1;
                                    printf("\tnew value of $$ is %f\n", f);
                                    $$ = f;
                                }
;

这是像 int [12][3] ciao; 这样的输入的输出

processing B -> int
processing B inside T
    value of B is 1.000000
processing C -> empty
    setting C to be equal to 1
processing C -> [3] C
    previous C value is 1.000000
    new value of $$ is 2.000000           
processing C -> [12] C                    
    previous C value is 2.000000          
    new value of $$ is 3.000000           
processing C inside T
processing T-> B C
    value of B is 1.000000
    value of C is 0.000000                (*)
processing D -> empty
processing D -> T var ; D
    value of T is 1.000000                (*)

如您所见，在用 * 标记的 C 缩减中值(value)丢失了，我希望它长大，如下所示

processing B -> int
processing B inside T
    value of B is 1.000000
processing C -> empty
    setting C to be equal to 1
processing C -> [3] C
    previous C value is 1.000000
    new value of $$ is 2.000000           
processing C -> [12] C                    
    previous C value is 2.000000          
    new value of $$ is 3.000000           
processing C inside T
processing T-> B C
    value of B is 1.000000
    value of C is 3.000000                
processing D -> empty
processing D -> T var ; D
    value of T is 4.000000                

感谢任何提示以及达到范围的解释和建议，我是否遗漏了什么？

Answer 1

T 的产生式如下，已大大简化。

T: B { /* Mid Rule Action (MRA) */ } C { $$ = $1 + $2; }

在 T 的最终操作中，$2 指的是 MRA，因为 MRA 被计算在生产条款中。 (实际上，MRA 被替换为具有空 RHS 的非终端。)所以 C 是 $3。

由于 MRA 实际上没有设置值，$2 有点不确定，但 0 也不太可能。

Bison 手册引用:

Using Mid-Rule Actions:

The mid-rule action itself counts as one of the components of the rule. This makes a difference when there is another action later in the same rule (and usually there is another at the end): you have to count the actions along with the symbols when working out which number n to use in $n.

Mid-Rule Action Translation:指出“中间规则 Action 实际上转化为常规规则和 Action ”，然后提供了一些生成的空规则示例(及其内部名称，有助于理解 bison 调试输出。)