strcpy using pointers

Question

I\'m trying to write strcpy on my own using pointers and I get an error during runtime.

void str_cpy(char **destination, const char *source) {
//    char *s1         


        

Answer 1

You should likely allocate some memory for that pointer before passing it off to a function that fills what it points to (which in this case, is NULL).

Example:

char *str = malloc(128);
if (str)
{
   str_cpy(&str, "String");
   free(str);
   str = NULL;
}

I advise not doing this without also providing target-buffer size information (i.e. if you're writing your own, then boundary-check the target buffer, otherwise your version has the same security flaws as strcpy() which are bad enough as it is).

Note: Unless you're planning on changing the address held by the pointer passed as the target, you need not use a double pointer either. The double pointer usage you have prevents the traditional strcpy() usage pattern of:

char str[128];
str_cpy(&str, "Hello"); // error. 

An array address cannot be passed as a pointer-to-pointer, so your code cannot fill a static array without an intermediate pointer:

char str[128];
char *p = str;
str_cpy(&p, "Hello");  //ok. passing address of pointer.

If this is not intentional (and I don't see why it could be unless you have ideas of internally emulating strdup() on a NULL pointer passage) You should address this.