%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,
%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,$1,$3); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",$2); }
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
{
presentlevel=presentlevel+1;
}
if($3=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi($1); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,$1);
$$=h->val;
}
| expr TOK_ADD expr {$$=$1+$2;}
| expr TOK_MUL expr {$$=$1*$2;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
I have declared the types for the id_val and int_val below the union as you can see. Still it is causing an error. Here are the errors that I am getting.
我已经在union下面声明了id_val和int_val的类型。它仍然导致错误。这是我得到的错误。
calc.y:46.44-45: $1 of `stmt' has no declared type
calc.y:51.43-44: $3 of `stmt' has no declared type
calc.y:65.47-48: $2 of `expr' has no declared type
calc.y:66.47-48: $2 of `expr' has no declared type
make: *** [calc] Error 1
Can somebody tell us why is it showing the error inspite of declaring the type.
有人可以告诉我们为什么它会显示错误,尽管声明了类型。
1 个解决方案
#1
1
Well, the error messages tell you pretty much exactly what is going on. For the first one, line 46 is:
好吧,错误消息告诉你几乎到底发生了什么。对于第一个,第46行是:
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
this is a rule for stmt, and the error tells you that $1 (which comes from TOK_LPARA) has no type. You can see that from its delaration:
这是stmt的规则,错误告诉你$ 1(来自TOK_LPARA)没有类型。你可以从它的delaration看到:
%left TOK_LPARA TOK_RPARA
If you want to be able to access a value from TOK_LPARA here, you need to give it a type, probably <id_val>. Then you'll have the problem that == will compare pointers, not pointed at strings. The other errors all indicate similar problems.
如果您希望能够在此处访问TOK_LPARA中的值,则需要为其指定一个类型,可能是
。然后你会遇到这样的问题:==会比较指针,而不是指向字符串。其他错误都表明了类似的问题。
You also have problems with not setting $$ in stmt actions (you've declared a type for stmt) causing them to have garbage values.
您还没有在stmt操作中设置$$(您已声明stmt的类型)导致它们具有垃圾值。
,); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",); }
| TOK_LPARA stmts TOK_RPARA { if(
%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,$1,$3); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",$2); }
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
{
presentlevel=presentlevel+1;
}
if($3=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi($1); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,$1);
$$=h->val;
}
| expr TOK_ADD expr {$$=$1+$2;}
| expr TOK_MUL expr {$$=$1*$2;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
I have declared the types for the id_val and int_val below the union as you can see. Still it is causing an error. Here are the errors that I am getting.
我已经在union下面声明了id_val和int_val的类型。它仍然导致错误。这是我得到的错误。
calc.y:46.44-45: $1 of `stmt' has no declared type
calc.y:51.43-44: $3 of `stmt' has no declared type
calc.y:65.47-48: $2 of `expr' has no declared type
calc.y:66.47-48: $2 of `expr' has no declared type
make: *** [calc] Error 1
Can somebody tell us why is it showing the error inspite of declaring the type.
有人可以告诉我们为什么它会显示错误,尽管声明了类型。
1 个解决方案
#1
1
Well, the error messages tell you pretty much exactly what is going on. For the first one, line 46 is:
好吧,错误消息告诉你几乎到底发生了什么。对于第一个,第46行是:
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
this is a rule for stmt, and the error tells you that $1 (which comes from TOK_LPARA) has no type. You can see that from its delaration:
这是stmt的规则,错误告诉你$ 1(来自TOK_LPARA)没有类型。你可以从它的delaration看到:
%left TOK_LPARA TOK_RPARA
If you want to be able to access a value from TOK_LPARA here, you need to give it a type, probably <id_val>. Then you'll have the problem that == will compare pointers, not pointed at strings. The other errors all indicate similar problems.
如果您希望能够在此处访问TOK_LPARA中的值,则需要为其指定一个类型,可能是
。然后你会遇到这样的问题:==会比较指针,而不是指向字符串。其他错误都表明了类似的问题。
You also have problems with not setting $$ in stmt actions (you've declared a type for stmt) causing them to have garbage values.
您还没有在stmt操作中设置$$(您已声明stmt的类型)导致它们具有垃圾值。
=="{")
{
presentlevel=presentlevel+1;
}
if(=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi(
%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,$1,$3); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",$2); }
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
{
presentlevel=presentlevel+1;
}
if($3=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi($1); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,$1);
$$=h->val;
}
| expr TOK_ADD expr {$$=$1+$2;}
| expr TOK_MUL expr {$$=$1*$2;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
I have declared the types for the id_val and int_val below the union as you can see. Still it is causing an error. Here are the errors that I am getting.
我已经在union下面声明了id_val和int_val的类型。它仍然导致错误。这是我得到的错误。
calc.y:46.44-45: $1 of `stmt' has no declared type
calc.y:51.43-44: $3 of `stmt' has no declared type
calc.y:65.47-48: $2 of `expr' has no declared type
calc.y:66.47-48: $2 of `expr' has no declared type
make: *** [calc] Error 1
Can somebody tell us why is it showing the error inspite of declaring the type.
有人可以告诉我们为什么它会显示错误,尽管声明了类型。
1 个解决方案
#1
1
Well, the error messages tell you pretty much exactly what is going on. For the first one, line 46 is:
好吧,错误消息告诉你几乎到底发生了什么。对于第一个,第46行是:
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
this is a rule for stmt, and the error tells you that $1 (which comes from TOK_LPARA) has no type. You can see that from its delaration:
这是stmt的规则,错误告诉你$ 1(来自TOK_LPARA)没有类型。你可以从它的delaration看到:
%left TOK_LPARA TOK_RPARA
If you want to be able to access a value from TOK_LPARA here, you need to give it a type, probably <id_val>. Then you'll have the problem that == will compare pointers, not pointed at strings. The other errors all indicate similar problems.
如果您希望能够在此处访问TOK_LPARA中的值,则需要为其指定一个类型,可能是
。然后你会遇到这样的问题:==会比较指针,而不是指向字符串。其他错误都表明了类似的问题。
You also have problems with not setting $$ in stmt actions (you've declared a type for stmt) causing them to have garbage values.
您还没有在stmt操作中设置$$(您已声明stmt的类型)导致它们具有垃圾值。
); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,
%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,$1,$3); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",$2); }
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
{
presentlevel=presentlevel+1;
}
if($3=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi($1); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,$1);
$$=h->val;
}
| expr TOK_ADD expr {$$=$1+$2;}
| expr TOK_MUL expr {$$=$1*$2;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
I have declared the types for the id_val and int_val below the union as you can see. Still it is causing an error. Here are the errors that I am getting.
我已经在union下面声明了id_val和int_val的类型。它仍然导致错误。这是我得到的错误。
calc.y:46.44-45: $1 of `stmt' has no declared type
calc.y:51.43-44: $3 of `stmt' has no declared type
calc.y:65.47-48: $2 of `expr' has no declared type
calc.y:66.47-48: $2 of `expr' has no declared type
make: *** [calc] Error 1
Can somebody tell us why is it showing the error inspite of declaring the type.
有人可以告诉我们为什么它会显示错误,尽管声明了类型。
1 个解决方案
#1
1
Well, the error messages tell you pretty much exactly what is going on. For the first one, line 46 is:
好吧,错误消息告诉你几乎到底发生了什么。对于第一个,第46行是:
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
this is a rule for stmt, and the error tells you that $1 (which comes from TOK_LPARA) has no type. You can see that from its delaration:
这是stmt的规则,错误告诉你$ 1(来自TOK_LPARA)没有类型。你可以从它的delaration看到:
%left TOK_LPARA TOK_RPARA
If you want to be able to access a value from TOK_LPARA here, you need to give it a type, probably <id_val>. Then you'll have the problem that == will compare pointers, not pointed at strings. The other errors all indicate similar problems.
如果您希望能够在此处访问TOK_LPARA中的值,则需要为其指定一个类型,可能是
。然后你会遇到这样的问题:==会比较指针,而不是指向字符串。其他错误都表明了类似的问题。
You also have problems with not setting $$ in stmt actions (you've declared a type for stmt) causing them to have garbage values.
您还没有在stmt操作中设置$$(您已声明stmt的类型)导致它们具有垃圾值。
);
$$=h->val;
}
| expr TOK_ADD expr {$$=
%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,$1,$3); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",$2); }
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
{
presentlevel=presentlevel+1;
}
if($3=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi($1); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,$1);
$$=h->val;
}
| expr TOK_ADD expr {$$=$1+$2;}
| expr TOK_MUL expr {$$=$1*$2;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
I have declared the types for the id_val and int_val below the union as you can see. Still it is causing an error. Here are the errors that I am getting.
我已经在union下面声明了id_val和int_val的类型。它仍然导致错误。这是我得到的错误。
calc.y:46.44-45: $1 of `stmt' has no declared type
calc.y:51.43-44: $3 of `stmt' has no declared type
calc.y:65.47-48: $2 of `expr' has no declared type
calc.y:66.47-48: $2 of `expr' has no declared type
make: *** [calc] Error 1
Can somebody tell us why is it showing the error inspite of declaring the type.
有人可以告诉我们为什么它会显示错误,尽管声明了类型。
1 个解决方案
#1
1
Well, the error messages tell you pretty much exactly what is going on. For the first one, line 46 is:
好吧,错误消息告诉你几乎到底发生了什么。对于第一个,第46行是:
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
this is a rule for stmt, and the error tells you that $1 (which comes from TOK_LPARA) has no type. You can see that from its delaration:
这是stmt的规则,错误告诉你$ 1(来自TOK_LPARA)没有类型。你可以从它的delaration看到:
%left TOK_LPARA TOK_RPARA
If you want to be able to access a value from TOK_LPARA here, you need to give it a type, probably <id_val>. Then you'll have the problem that == will compare pointers, not pointed at strings. The other errors all indicate similar problems.
如果您希望能够在此处访问TOK_LPARA中的值,则需要为其指定一个类型,可能是
。然后你会遇到这样的问题:==会比较指针,而不是指向字符串。其他错误都表明了类似的问题。
You also have problems with not setting $$ in stmt actions (you've declared a type for stmt) causing them to have garbage values.
您还没有在stmt操作中设置$$(您已声明stmt的类型)导致它们具有垃圾值。
+;}
| expr TOK_MUL expr {$$=
%{
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include "mycalc.h"
extern int int_num;
extern char* yytext;
%}
%token TOK_NUM TOK_ID TOK_SEMICOLON TOK_VAR TOK_EQ TOK_PRINTLN TOK_LPARA TOK_RPARA TOK_ADD TOK_MUL
%union
{
int int_val;
char *id_val;
}
%type <id_val> expr TOK_ID
%type <int_val> stmt TOK_NUM
%left TOK_LPARA TOK_RPARA
%left TOK_MUL
%left TOK_ADD
%%
prog:
stmts { startit(); }
;
stmts:
| stmt TOK_SEMICOLON stmts
;
stmt:
TOK_VAR TOK_ID { defvar(presentlevel,yylval.id_val,0); }
| TOK_ID TOK_EQ expr { assignvar(presentlevel,$1,$3); }
| TOK_PRINTLN TOK_ID { printf("the value of id %d",$2); }
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
{
presentlevel=presentlevel+1;
}
if($3=="}")
{
if(presentlevel>1)
{
presentlevel=presentlevel-1;
}
} };
expr:
TOK_NUM { $$=atoi($1); }
| TOK_ID { myvar *h ;
h=getvar(presentlevel,$1);
$$=h->val;
}
| expr TOK_ADD expr {$$=$1+$2;}
| expr TOK_MUL expr {$$=$1*$2;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
I have declared the types for the id_val and int_val below the union as you can see. Still it is causing an error. Here are the errors that I am getting.
我已经在union下面声明了id_val和int_val的类型。它仍然导致错误。这是我得到的错误。
calc.y:46.44-45: $1 of `stmt' has no declared type
calc.y:51.43-44: $3 of `stmt' has no declared type
calc.y:65.47-48: $2 of `expr' has no declared type
calc.y:66.47-48: $2 of `expr' has no declared type
make: *** [calc] Error 1
Can somebody tell us why is it showing the error inspite of declaring the type.
有人可以告诉我们为什么它会显示错误,尽管声明了类型。
1 个解决方案
#1
1
Well, the error messages tell you pretty much exactly what is going on. For the first one, line 46 is:
好吧,错误消息告诉你几乎到底发生了什么。对于第一个,第46行是:
| TOK_LPARA stmts TOK_RPARA { if($1=="{")
this is a rule for stmt, and the error tells you that $1 (which comes from TOK_LPARA) has no type. You can see that from its delaration:
这是stmt的规则,错误告诉你$ 1(来自TOK_LPARA)没有类型。你可以从它的delaration看到:
%left TOK_LPARA TOK_RPARA
If you want to be able to access a value from TOK_LPARA here, you need to give it a type, probably <id_val>. Then you'll have the problem that == will compare pointers, not pointed at strings. The other errors all indicate similar problems.
如果您希望能够在此处访问TOK_LPARA中的值,则需要为其指定一个类型,可能是
。然后你会遇到这样的问题:==会比较指针,而不是指向字符串。其他错误都表明了类似的问题。
You also have problems with not setting $$ in stmt actions (you've declared a type for stmt) causing them to have garbage values.
您还没有在stmt操作中设置$$(您已声明stmt的类型)导致它们具有垃圾值。
*;}
;
%%
int yyerror(char *s,int x)
{
printf("Syntax Error at %d",line_num);
return 0;
}
int main()
{
startit();
presentlevel=1;
yyparse();
return 0;
}
%{
#include<stdio.h>
#include<string.h>
#inclu
阅读背景：
野牛：'stmt'的1美元没有声明的类型
来源：互联网
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