我正在按照 [编译器构造、原理和实践] 制作 C 编译器。 我几乎已经使用 yacc 制作了解析器，但是 yacc 中的一条规则出错了。
return_stmt : RETURN SEMI { $$ = newStmtNode(ReturnK);}
| RETURN expression SEMI
{ $$ = newStmtNode(ReturnK);
$$->child[0] = $1;
}
当使用“RETURN 表达式 SEMI”规则解析 return 语句时，$1 的值为零。整个来源如下 %%/* C-语法 */
program : declaration_list
{ savedTree = $1; }
;
declaration_list : declaration_list declaration
{ YYSTYPE t = $1;
if (t != NULL)
{ while (t->sibling != NULL)
t = t->sibling;
t->sibling = $2;
$$ = $1; }
else $$ = $2;
}
| declaration { $$ =$1; }
;
declaration : var_declaration { $$ = $1; }
| fun_declaration { $$ = $1; }
;
var_declaration : type_specifier ID SEMI
{ $$ = newExpNode(VarK);
$$->type = (ExpType)$1;
$$->attr.name = nameStackPop();
$$->lineno = savedLineNo;
}
| type_specifier ID LBRACE NUM RBRACE SEMI
{ $$ = newExpNode(VarArrayK);
$$->type = (ExpType)$1;
$$->attr.name = nameStackPop();
$$->lineno = savedLineNo;
$$->child[0] = $4;
}
;
type_specifier : INT { $$ = Integer; }
| VOID { $$ = Void; }
;
fun_declaration : type_specifier ID LPAREN params RPAREN compound_stmt
{ $$ = newStmtNode(FunctionK);
$$->type = (ExpType)$1;
$$->attr.name = nameStackPop();
$$->lineno = savedLineNo;
$$->child[0] = $4;
$$->child[1] = $6;
}
;
params : param_list { $$ = $1; }
| VOID
{ $$ = newExpNode(SingleParamK);
$$->type = Void;
}
;
param_list : param_list COMMA param
{ YYSTYPE t = $1;
if (t != NULL)
{ while (t->sibling != NULL)
t = t->sibling;
t->sibling = $3;
$$ = $1; }
else $$ = $3;
}
| param { $$ = $1; }
;
param : type_specifier ID
{ $$ = newExpNode(SingleParamK);
$$->type = (ExpType)$1;
$$->attr.name = nameStackPop();
}
| type_specifier ID LBRACE RBRACE
{ $$ = newExpNode(ArrayParamK);
$$->type = (ExpType)$1;
$$->attr.name = nameStackPop();
$$->lineno = savedLineNo;
}
;
compound_stmt : LCURLY local_declarations statement_list RCURLY
{ $$ = newStmtNode(CompoundK);
$$->child[0] = $2;
$$->child[1] = $3;
}
;
local_declarations : local_declarations var_declaration
{ YYSTYPE t = $1;
if (t != NULL)
{ while (t->sibling != NULL)
t = t->sibling;
t->sibling = $2;
$$ = $1; }
else $$ = $2;
}
| empty { $$ = $1; }
;
statement_list : statement_list statement
{ YYSTYPE t = $1;
if (t != NULL)
{ while (t->sibling != NULL)
t = t->sibling;
t->sibling = $2;
$$ = $1; }
else $$ = $2;
}
| empty { $$ = $1; }
;
statement : expression_stmt { $$ = $1; }
| compound_stmt { $$ = $1; }
| selection_stmt { $$ = $1; }
| iteration_stmt { $$ = $1; }
| return_stmt { $$ = $1; }
;
expression_stmt : expression SEMI { $$ = $1;}
| SEMI { $$ = NULL; }
;
selection_stmt : IF LPAREN expression RPAREN statement
{ $$ = newStmtNode(IfK);
$$->child[0] = $3;
$$->child[1] = $5;
$$->attr.withElse = FALSE;
}
| IF LPAREN expression RPAREN statement ELSE statement
{ $$ = newStmtNode(IfK);
$$->child[0] = $3;
$$->child[1] = $5;
$$->child[2] = $7;
$$->attr.withElse = TRUE;
}
;
iteration_stmt : WHILE LPAREN expression RPAREN statement
{ $$ = newStmtNode(WhileK);
$$->child[0] = $3;
$$->child[1] = $5;
}
;
return_stmt : RETURN SEMI { $$ = newStmtNode(ReturnK);}
| RETURN expression SEMI
{ $$ = newStmtNode(ReturnK);
$$->child[0] = $1;
}
;
expression : var ASSIGN expression
{ $$ = newExpNode(AssignK);
$$->type = Integer;
$$->child[0] = $1;
$$->child[1] = $3;
}
| simple_expression { $$ = $1; }
;
var : ID
{ $$ = newExpNode(IdK);
$$->type = Integer;
$$->attr.name = nameStackPop();
}
| ID LBRACE expression RBRACE
{ $$ = newExpNode(IdK);
$$->attr.name = nameStackPop();
$$->child[0] = $3;
}
;
simple_expression : additive_expression relop additive_expression
{ $$ = newExpNode(OpK);
$$->type = Integer;
$$->child[0] = $1;
$$->child[1] = $3;
$$->attr.op = $2;
}
| additive_expression
{ $$ = $1; }
;
relop : LE { $$ = LE; }
| LT { $$ = LT; }
| GT { $$ = GT; }
| GE { $$ = GE; }
| EQ { $$ = EQ; }
| NE { $$ = NE; }
;
additive_expression : additive_expression addop term
{ $$ = newExpNode(OpK);
$$->type = Integer;
$$->child[0] = $1;
$$->child[1] = $3;
$$->attr.op = $2;
}
| term { $$ = $1; }
;
addop : PLUS { $$ = PLUS; }
| MINUS { $$ = MINUS; }
;
term : term mulop factor
{ $$ = newExpNode(OpK);
$$->type = Integer;
$$->child[0] = $1;
$$->child[1] = $3;
$$->attr.op = $2;
}
| factor { $$ = $1; }
;
mulop : TIMES { $$ = TIMES; }
| OVER { $$ = OVER; }
;
factor : LPAREN expression RPAREN
{ $$ = $2; }
| var { $$ = $1; }
| call { $$ = $1; }
| NUM
{ $$ = newExpNode(ConstK);
$$->type = Integer;
$$->attr.val = atoi(tokenString);
}
;
call : ID LPAREN args RPAREN
{ $$ = newExpNode(CallK);
$$->attr.name = nameStackPop();
$$->child[0] = $3;
$$->lineno = savedLineNo;
}
;
args : arg_list
{ $$ = $1; }
| empty
{ $$ = $1; }
;
arg_list : arg_list COMMA expression
{ YYSTYPE t = $1;
if (t != NULL)
{ while (t->sibling != NULL)
t = t->sibling;
t->sibling = $3;
$$ = $1; }
else $$ = $3;
}
| expression
{ $$ = $1; }
;
empty : { $$ = NULL;}
;
%%
当我使用 printf 进行调试时，我确认规则“RETURN 表达式 SEMI”中的所有其他非终结符都具有正确的值(树节点的指针)。这个 yacc 文件有什么问题？
最佳答案
$1 是return 标记。 $2 应该是表达式。
关于c - yacc 从非终端获取零值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26561084/