我必须使用flex和bison实现表达式树的解析器(比如“(a> b)AND(c< = d)”),但我无法解决类型错误...
在parser.y.c文件的g ++编译期间发生致命错误(由此命令生成:“bison -o parser.y.c -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:警告,larèglenepeutêtrepairée[英文:“规则无法匹配”]
我希望有人可以帮助我!
这里是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 { $$=$2; }
;
%%
void parse_string(const std::string & str)
{
yy_scan_string(str.c_str());
yyparse();
}
然后是parser.lex文件:
%{
#define YYSTYPE Node*
#include <cstdlib>
#include "BooleanNode.h"
#include "AttributeNode.h"
#include "parser.y.h"
extern "C++"
{
int yylex(void);
}
%}
%option noyywrap
blancs [ \t]+
ident [a-zA-Z_]{1}[a-zA-Z0-9_]*
%%
{ident} { return(IDENT); }
"<" return(LT);
">" return(GT);
"<=" return(LE);
">=" return(GE);
"=" return(EQ);
"!=" return(NE);
"AND" return(AND);
"OR" return(OR);
"(" return(LEFT_PARENTHESIS);
")" return(RIGHT_PARENTHESIS);
"\n" return(FIN);
最后是Node.h文件:
#ifndef _NODE_H_
#define _NODE_H_
#include <string>
#include <iostream>
class Node
{
public:
enum E_op
{
AND = 0,
OR,
LT,
GT,
LE,
GE,
EQ,
NE
};
Node(const std::string & op)
{
_op = op;
}
Node(const std::string & op, const std::string & left, const std::string & right)
{
_op = op;
}
Node(const std::string & op, Node * left, Node * right)
{
_op = op;
}
virtual ~Node()
{
}
virtual void print() {}
protected:
std::string _op;
};
#endif
更新
感谢Jonathan Leffler和其他一些更正(char *而不是%union中的std :: string),编译顺利但结果不是我的预期。 使用“foo&lt; bar”表达式,执行“IDENT LT IDENT”指令,但$ 1和$ 3的值为NULL ...
**新更新**
我通过拆分Expression指令来纠正错误:
Expression:
id LT id { $$ = 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); }
答案 0 :(得分:1)
问题在于您已通过Node *
定义声明了Yacc堆栈包含#define YYSTYPE Node *
个元素,但您的%union
,%token
和%type
声明说联盟中有StrVal
和NodeVal
类型。
IIRC,您只在不使用YYSTYPE
时使用%union
。删除该行应解决其他问题。