赋值从没有强制转换的指针中获取整数

Question

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;
}

In the above code, the function addstring(..) shoes this error Assignment makes integer from a pointer without a cast . I know I'm taking the value of a pointer and putting it in integer, but how may I do it to resolve this error?

Answer 1

Even after changing *dest to dest , your function addstring is not works properly.. Simply try like this

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;
}

Answer 2

You have done

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

instead of

dest =malloc(buflen + len + 1);

Your program saying warning to me for this line

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

sizeof() return type is not long unsigned int.

So use %d or %u or %zu as a access specifier in printf() statement.

Answer 3

change

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

to

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

EDIT: As @POW said, you need not cast the result of malloc

Answer 4

There are multiple issue in your code. Please check the below code

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;
}