gcc错误:对'itoa'的未定义引用

Question

如果我包含stdlib.h,那么也无法识别itoa().我的代码:

%{
#include "stdlib.h"
#include <stdio.h>
#include <math.h>
int yylex(void);
char p[10]="t",n1[10];
int n ='0';

%}
%union
{
char *dval;
}
%token ID
%left '+' '-'
%left '*' '/'
%nonassoc UMINUS
%type <dval> S
%type <dval> E
%%
S : ID '=' E {printf(" x = %sn",$$);}
;
E : ID {}
| E '+' E {n++;itoa(n,n1,10);printf(" %s = %s + %s ",p,$1,$3);strcpy($$,p);strcat($$,n1);}
| E '-' E {n++;itoa(n,n1,10);printf(" %s = %s – %s ",p,$1,$3);strcpy($$,p);strcat($$,n1);}
| E '*' E {n++;itoa(n,n1,10);printf(" %s = %s * %s ",p,$1,$3);strcpy($$,p);strcat($$,n1);}
| E '/' E {n++;itoa(n,n1,10);printf(" %s = %s / %s ",p,$1,$3);strcpy($$,p);strcat($$,n1);}
;
%%

main()
{
yyparse();
}

int yyerror (char *s)


{


}

运行我得到的代码后:

gcc lex.yy.c y.tab.c -ll
12.y: In function ‘yyparse’:
12.y:24: warning: incompatible implicit declaration of built-in function ‘strcpy’
12.y:24: warning: incompatible implicit declaration of built-in function ‘strcat’
12.y:25: warning: incompatible implicit declaration of built-in function ‘strcpy’
12.y:25: warning: incompatible implicit declaration of built-in function ‘strcat’
12.y:26: warning: incompatible implicit declaration of built-in function ‘strcpy’
12.y:26: warning: incompatible implicit declaration of built-in function ‘strcat’
12.y:27: warning: incompatible implicit declaration of built-in function ‘strcpy’
12.y:27: warning: incompatible implicit declaration of built-in function ‘strcat’
/tmp/ccl0kjje.o: In function `yyparse':
y.tab.c:(.text+0x33d): undefined reference to `itoa'
y.tab.c:(.text+0x3bc): undefined reference to `itoa'
y.tab.c:(.text+0x43b): undefined reference to `itoa'
y.tab.c:(.text+0x4b7): undefined reference to `itoa'

我哪里错了？为什么它找不到itoa的引用？我也尝试使用<>括号进行itoa.

Answer 1

itoa是一些非标准函数,由一些编译器支持.由于错误,编译器不支持它.你最好的选择是使用snprintf().

Answer 2

我使用sprintf以下方式：

#include <stdio.h>

char *my_itoa(int num, char *str)
{
        if(str == NULL)
        {
                return NULL;
        }
        sprintf(str, "%d", num);
        return str;
}

int main()
{
        int num = 2016;
        char str[20];
        if(my_itoa(num, str) != NULL)
        {
                printf("%s\n", str);
        }
}

我希望这会节省某人的时间;)