Question

yacc的语法如下：

Successfully fetched weather data!

现在，当我在.y文件上运行bison时，我收到以下错误：

18. Stmt_list : Stmt | Stmt_list ';' Stmt;
19. Stmt : Variable assignop Expression;
20. Variable : id | id'['Expression']';
21. Expression : Simple_expression | Simple_expression relop Simple_expression;
22. Simple_expression : Term | Simple_expression addop Term;
23. Term : Factor | Term mulop Factor;
24. Factor : id | num | '('Expression')' | id'['Expression']' | not Factor;

我是野牛的完整菜鸟。我不知道如何在这里解决问题。

/ **输入和输出文件** /

输入：

subc.y: conflicts: 8 shift/reduce, 26 reduce/reduce
subc.y:21.21-45: warning: rule useless in parser due to conflicts: expr: simp_ex
pr relop simp_expr
subc.y:22.20-39: warning: rule useless in parser due to conflicts: simp_expr: si
mp_expr addop term
subc.y:23.17-33: warning: rule useless in parser due to conflicts: term: term mu
lop factor
subc.y:24.40-49: warning: rule useless in parser due to conflicts: factor: not f
actor

输出：

%{
#include<stdio.h>
#include "lex.yy.c"
    void yyerror (char *s);
%}

        /* Yacc definitions */
%start stmt_list
%token id
%token num
%token relop
%token mulop
%token addop
%token assignop
%token not
%%
/* descriptions of expected inputs corresponding actions (in C) */
stmt_list   : stmt| stmt_list ';' stmt  { printf("SUCCESS"); };
stmt  : var assignop expr ;
var   : id| id'['expr']' ;
expr  : simp_expr | simp_expr relop simp_expr;
simp_expr  : term| simp_expr addop term;
term  : factor| term mulop factor;
factor  : id| num| expr| id'['expr']'| not factor;
%%

void yyerror (char *s) {printf (stderr, "%s\n", s);}

Answer 1

这很简单，构造：

id '[' expr ']' | id

可以是var和factor。

语法含糊不清，这就是野牛告诉你的......

解决方法是将其从factor中删除，然后在其位置插入var。

语法中也有一个循环：

expr -> simp_exp -> term -> factor -> expr

从expr移除factor以消除该循环，或反映原始语法中的内容，或许应该是：

  '(' expr ')'

（你忘记了表达式的括号）

转移/减少野牛的错误

1 个答案: