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我必须使用 flex 和 bison 实现表达式树的解析器（如“（a > b）AND（c <= d）”），但我无法解决类型错误......

致命错误发生在 parser.yc 文件的 g++ 编译期间（由以下命令生成：“bison -o parser.yc -d parser.y”）：

parser.y:54:36: erreur: request for member ‘nodeVal’ in ‘*(yyvsp + -8u)’, which is of pointer type ‘Node*’ (maybe you meant to use ‘->’ ?)
parser.y:58:14: erreur: request for member ‘nodeVal’ in ‘yyval’, which is of pointer type ‘Node*’ (maybe you meant to use ‘->’ ?)
parser.y:58:55: erreur: request for member ‘strVal’ in ‘*(yyvsp + -16u)’, which is of pointer type ‘Node*’ (maybe you meant to use ‘->’ ?)
parser.y:58:82: erreur: request for member ‘strVal’ in ‘* yyvsp’, which is of pointer type ‘Node*’ (maybe you meant to use ‘->’ ?)
parser.y:59:14: erreur: request for member ‘nodeVal’ in ‘yyval’, which is of pointer type ‘Node*’ (maybe you meant to use ‘->’ ?)
parser.y:59:55: erreur: request for member ‘strVal’ in ‘*(yyvsp + -16u)’, which is of pointer type ‘Node*’ (maybe you meant to use ‘->’ ?)

还有一个我不明白的警告：parser.lex:35: warning, la règle ne peut être pairée [英语：“规则无法匹配”]

我希望有一个人可以帮助我 ！

在这里，parser.y 文件：

%{

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>

#include "Node.h"
#include "parser.lex.h"

#define YYSTYPE Node*

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

extern "C++"
{
    int yyparse(void);
    int yylex(void);
    Node * rootNode;
}

%}

%union {
    Node * nodeVal;
    char * strVal;
}

%token <strVal> IDENT
%token <strVal> LT GT LE GE EQ NE
%token <strVal> AND OR
%token <strVal> LEFT_PARENTHESIS RIGHT_PARENTHESIS
%token FIN

%left   LT GT LE GE EQ NE
%left   AND OR

%type<nodeVal> Expression

%start Input
%%

Input:
          /* Vide */
        | Input Ligne
        ;

Ligne:
          FIN
        | Expression FIN                { rootNode = $1; }
        ;

Expression:
          IDENT LT IDENT  { $$=new Node("<", $1, $3); }
        | IDENT GT IDENT  { $$=new Node(">", $1, $3); }
        | IDENT LE IDENT  { $$=new Node("<=", $1, $3); }
        | IDENT GE IDENT  { $$=new Node(">=", $1, $3); }
        | IDENT EQ IDENT  { $$=new Node("=", $1, $3); }
        | IDENT NE IDENT  { $$=new Node("!=", $1, $3); }
        | Expression AND Expression  { $$=new Node("AND", $1, $3); }
        | Expression OR Expression  { $$=new Node("OR", $1, $3); }
        | LEFT_PARENTHESIS Expression RIGHT_PARENTHESIS        { $$=$2; }
        ;

%%

void parse_string(const std::string & str)
{
    yy_scan_string(str.c_str());
    yyparse();
}

然后是 parser.lex 文件：

%{

#define YYSTYPE Node*

#include <cstdlib>

#include "BooleanNode.h"
#include "AttributeNode.h"
#include "parser.y.h"

extern "C++"
{
    int yylex(void);
}

%}

%option noyywrap

blancs          [ \t]+

ident           [a-zA-Z_]{1}[a-zA-Z0-9_]*

%%

{ident}         { return(IDENT); }

"<"    return(LT);
">"    return(GT);
"<="   return(LE);
">="   return(GE);
"="    return(EQ);
"!="   return(NE);

"AND"  return(AND);
"OR"   return(OR);

"("    return(LEFT_PARENTHESIS);
")"    return(RIGHT_PARENTHESIS);

"\n"  return(FIN);

最后是 Node.h 文件：

#ifndef _NODE_H_
#define _NODE_H_

#include <string>
#include <iostream>


class Node
{
public:

    enum E_op
    {
        AND = 0,
        OR,
        LT,
        GT,
        LE,
        GE,
        EQ,
        NE
    };

    Node(const std::string & op)
    {
        _op = op;
    }

    Node(const std::string & op, const std::string & left, const std::string & right)
    {
        _op = op;
    }

    Node(const std::string & op, Node * left, Node * right)
    {
        _op = op;
    }

    virtual ~Node()
    {

    }

    virtual void print() {}

protected:

    std::string _op;
};


#endif

更新

感谢 Jonathan Leffler 和其他一些更正（char* 而不是 %union 中的 std::string），编译进行得很顺利，但结果不是我所期望的。使用“foo < bar”表达式，执行“IDENT LT IDENT”指令，但 $1 和 $3 的值为 NULL...

** 新更新 **

我通过拆分 Expression 指令纠正了错误：

Expression:
          id LT id  { $$ = new Node("<", $1, $3); }
        | id GT id  { $$ = new Node(">", $1, $3); }
        | id LE id  { $$ = new Node("<=", $1, $3); }
        | id GE id  { $$ = new Node(">=", $1, $3); }
        | id EQ id  { $$ = new Node("=", $1, $3); }
        | id NE id  { $$ = new Node("!=", $1, $3); }
        | Expression AND Expression  { $$ = new Node("AND", $1, $3); }
        | Expression OR Expression   { $$ = new Node("OR", $1, $3); }
        | LEFT_PARENTHESIS Expression RIGHT_PARENTHESIS   { $$ = $2; }
        ;

id:
          IDENT     { $$ = strdup(yytext); }
4

1 回答 1

1

问题是您已经通过define声明Yacc 堆栈包含Node *元素，但是您的,和声明说在联合中有和类型。#define YYSTYPE Node *%union%token%typeStrValNodeVal

IIRC，您仅YYSTYPE在不使用时使用%union. 删除该行应该可以解决其他问题。

于 2013-07-19T14:44:29.983 回答