赋值使指针无整型指针

-1

int main(int argc, char *argv[]) { 
    if(argc!=3) { 
     printf("You must pass exactly three para \n"); 
     return 0; 
    } 

    char *buffer = argv[1]; 
    //printf("The length of the buffer string is %d\n",buflen); 
    char *mystring = argv[2]; 
    //printf("The length of the user string is %d\n",len); 
    addstring(buffer, mystring); 
    return 0; 
} 

int addstring(char *buffer, char *mystring) 
{ 
    int buflen = strlen(buffer); 
    int len = strlen(mystring); 
    char *dest; 
    *dest = (char *)malloc(buflen + len + 1); 
    printf("The size of destination is %lu\n",sizeof(dest)); 

    dest = strcpy(dest,buffer); 
    dest = (dest + buflen); 
    dest = strcpy(dest,mystring); 
    printf("The final string is %p",dest); 
    return 0; 
}

在上面的代码中，函数addstring(..)鞋这个错误Assignment makes integer from a pointer without a cast。我知道我正在接受一个指针的值并将其放入整数，但我该如何解决这个错误？赋值使指针无整型指针

来源

2014-10-31 lolly pop

你已经做了'* DEST =（的char *）malloc的（buflen + LEN + 1）;'而不是'DEST =的malloc（buflen + LEN + 1）;' – Chandru 2014-10-31 07:16:46

你需要'#include '和'#include ' – 2014-10-31 08:12:27

变化

char *dest; 
*dest = (char *)malloc(buflen + len + 1);

到

char *dest; 
dest = (char *)malloc(buflen + len + 1);

编辑：由于@POW说，你不用投malloc的结果

来源

2014-10-31 07:17:55

不要施加'malloc'结果 – P0W 2014-10-31 07:18:48

这与已发布的评论有何不同？如果你要发表评论作为答案，至少花时间解释你的答案背后的推理。 – 2014-10-31 07:18:53

@ DavidC.Rankin我发布我的答案时没有看到评论。相同的答案可能会在几秒钟内发布。 – 2014-10-31 07:28:03

即使改变*dest到dest后，你的函数addstring是不正常工作..只需尝试像这样

int addstring(char *buffer, char *mystring) 
{ 
int buflen = strlen(buffer); 
int len = strlen(mystring); 
char *dest; 
dest = (char *)malloc(buflen + len + 1); 
printf("The size of destination is %d\n",sizeof(dest)); 

strcpy(dest,buffer); 
strcat(dest,mystring); 
printf("The final string is %s\n",dest); 
return 0; 
}

来源

2014-10-31 07:25:41

你做

*dest = (char *)malloc(buflen + len + 1);

，而不是

dest =malloc(buflen + len + 1);

你的程序警告说，对我此行

printf("The size of destination is %lu\n",sizeof(dest));

sizeof()返回类型不长unsigned int类型。

因此，使用%d或%u或%zu作为printf()语句中的访问说明符。

来源

2014-10-31 07:27:37 Chandru

'sizeof'返回'size_t'类型，所以转换说明符应该是'％zu'。或者如果它不被编译器支持，那么使用显式转换'printf（“％u \ n”，（unsigned）sizeof（dest））;' – user694733 2014-10-31 07:37:51

感谢您的评论，我纠正了我的错误 – Chandru 2014-10-31 07:43:18

代码中存在多个问题。请检查下面的代码

int addstring(char *buffer, char *mystring) 
{ 
    int buflen = strlen(buffer); 
    int len = strlen(mystring); 
    char *dest; 
    /* No need to type-cast the malloc() */ 
    dest = malloc(buflen + len + 1); /* *dest holds the value, dest holds the address */ 
    printf("The size of destination is %lu\n",sizeof(dest)); 

    strcpy(dest,buffer); 
    strcpy(dest+buflen,mystring);/* copy the second string to dest after buffer is copied */ 
    printf("The final string is %s\n",dest); /*To print a string use %s, %p is to print pointer*/ 
    return 0; 
}

来源

2014-10-31 07:51:49 Gopi

赋值使指针无整型指针

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