strcpy vs strdup

I read that strcpy is for copying a string, and strdup returns a pointer to a new string to duplicate the string.

Could you please explain what cases do you prefer to use strcpy and what cases do you prefer to use strdup?

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strdup allocates memory for the new string on the heap, while using strcpy (or its safer strncpy varient) I can copy a string to a pre allocated memory on either the heap or the stack.

strcpy(ptr2, ptr1) is equivalent to while(*ptr2++ = *ptr1++)

where as strdup is equivalent to

ptr2 = malloc(strlen(ptr1)+1);
strcpy(ptr2,ptr1);

(memcpy version might be more efficient)

So if you want the string which you have copied to be used in another function (as it is created in heap section) you can use strdup, else strcpy is enough.

The functions strcpy and strncpy are part of the C standard library and operate on existing memory. That is, you must provide the memory into which the functions copy the string data, and as a corollary, you must have your own means of finding out how much memory you need.

By constrast, strdup is a Posix function, and it performs dynamic memory allocation for you. It returns a pointer to newly allocated memory into which it has copied the string. But you are now responsible for this memory and must eventually free it.

That makes strdup one of the "hidden malloc" convenience functions, and that's presumably also why it is not part of the standard library. As long as you use the standard library, you know that you must call one free for every malloc/calloc. But functions such as strdup introduce a hidden malloc, and you must treat it the same as a malloc for the purpose of memory management. (Another such hidden allocation functions is GCC's abi::__cxa_demangle().) Beware!

char *strdup(char *pszSrch);

strdup will allocate storage the size of the original string. If storage allocation is successful, the original string is copied to the duplicate string.

strdupd return NULL on failure. If memory is not allocated, copy fails strdup return NULL.

In the accepted answer, the implementation of strdup is presented as:

ptr2 = malloc(strlen(ptr1)+1);
strcpy(ptr2,ptr1);

However, that is somewhat sub-optimal because both strlen and strcpy need to find the length of the string by checking if each character is a \0.

Using memcpy should be more efficient:

char *strdup(const char *src) {
size_t len = strlen(src) + 1;
char *s = malloc(len);
if (s == NULL)
return NULL;
return (char *)memcpy(s, src, len);
}

for *memcpy another implementation is possible without incremental of dest and src :

while ( n-- ) { dest[n] = src[n] }