賦值使指針無整型指針

-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; 
}

在上面的代碼中，函數addstring(..)鞋這個錯誤Assignment makes integer from a pointer without a cast。我知道我正在接受一個指針的值並將其放入整數，但我該如何解決這個錯誤？賦值使指針無整型指針

來源

2014-10-31 lolly pop

你已經做了'* 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

變化

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

到

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

編輯：由於@POW說，你不用投malloc的結果

來源

2014-10-31 07:17:55

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

這與已發佈的評論有何不同？如果你要發表評論作爲答案，至少花時間解釋你的答案背後的推理。 – 2014-10-31 07:18:53

@ DavidC.Rankin我發佈我的答案時沒有看到評論。相同的答案可能會在幾秒鐘內發佈。 – 2014-10-31 07:28:03

即使改變*dest到dest後，你的函數addstring是不正常工作..只需嘗試像這樣

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; 
}

來源

2014-10-31 07:25:41

你做

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

，而不是

dest =malloc(buflen + len + 1);

你的程序警告說，對我此行

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

sizeof()返回類型不長unsigned int類型。

因此，使用%d或%u或%zu作爲printf()語句中的訪問說明符。

來源

2014-10-31 07:27:37 Chandru

'sizeof'返回'size_t'類型，所以轉換說明符應該是'％zu'。或者如果它不被編譯器支持，那麼使用顯式轉換'printf（「％u \ n」，（unsigned）sizeof（dest））;' – user694733 2014-10-31 07:37:51

感謝您的評論，我糾正了我的錯誤 – Chandru 2014-10-31 07:43:18

代碼中存在多個問題。請檢查下面的代碼

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; 
}

來源

2014-10-31 07:51:49 Gopi

賦值使指針無整型指針

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