Returning a pointer from a function

The C11 standards state that the strcpy function function must return a copy of its (original) destination parameter.

In both cases, we are returning a copy of the (original) dest parameter - that is, we are returning a pointer as the function's value.

We say that "the function's return-type is a pointer".

char *strcpy_array( char dest[], char src[] )
{
    int   i = 0;

    while( src[i] != '\0' ) {
	dest[ i ] = src[ i ];
	++i;
    }
    dest[ i ] = '\0';

    return dest;     // returns original destination parameter
}

char *strcpy_pointer( char *dest, char *src ) // two pointer parameters
{
    char *origdest = dest;  // take a copy of the dest parameter

    while( *src != '\0' ) {
        *dest = *src;       // copy one character from src to dest
        ++src;
        ++dest;
    }
    *dest = '\0';

    return origdest; // returns copy of original destination parameter
}

• in the array version, the function's return type is a pointer to a char. This further demonstrates the equivalence between array names (here, dest) and a pointer to the first element of that array.

• in the pointer version, we move the dest parameter after we have copied each character, and thus we must first save and then return a copy of the parameter's original value.

• if very careful, we could reduce the whole loop to the statement  while((*dest++ = *src++));
  But don't.