Is it possible to copy a string using C’s strcpy without first assigning memory to a char *character_array?

Basically, is something like this possible:

strcpy(char *my_string, another_string);

where my_string is an initialised string with the same length and content as another_string.

>Solution :

strcpy never allocates memory. The destination must be a pointer to enough, writable, validly-allocated memory, and you are responsible for making sure that the destination memory is validly allocated, long enough, and writable.

So these are all okay:

char *another_string = "hello";
char my_string_1[20];
strcpy(my_string_1, another_string);     /* valid (plenty of space) */

char my_string_2[6];
strcpy(my_string_2, another_string);     /* valid (just enough space) */

char my_string_3[] = "world";
strcpy(my_string_2, another_string);     /* valid (just enough space) */

char buf[6];
char *my_string_4 = buf;
strcpy(my_string_4, another_string);     /* valid (via intermediate pointer) */

char *my_string_5 = malloc(6);
strcpy(my_string_5, another_string);     /* valid (assuming malloc succeeds) */

char *my_string_6 = malloc(strlen(another_string) + 1);
strcpy(my_string_6, another_string);     /* valid (assuming malloc succeeds) */

But these are not valid:

char my_string_7[5];
strcpy(my_string_7, another_string);     /* INVALID (not enough space) */

char *my_string_8 = "world";
strcpy(my_string_8, another_string);     /* INVALID (destination not writable) */

char *my_string_9;
strcpy(my_string_9, another_string);     /* INVALID (destination not initialized) */

char *exceptionally_bad_allocator()
{
    char local_buffer[20];
    return local_buffer;
}

char *my_string_10 = exceptionally_bad_allocator();
strcpy(my_string_10, another_string);    /* INVALID (destination not validly allocated) */