C - fastest method to swap two memory blocks of equal size?

Question

What is the fastest way to swap two non-overlapping memory areas of equal size? Say, I need to swap (t_Some *a) with (t_Some *b). Considering space-tim

Answer 1

#include 
#include 

static void swap_elements_of_array(void* base, size_t size_of_element, int a, int b);
static void swap_elements_of_array(void* base, size_t size_of_element, int a, int b)
{
union {
    int i; /* force alignment */
    char zzz[size_of_element] ; /* VLA */
    } swap;
memcpy (swap.zzz, (char*)base + a * size_of_element,size_of_element);
memcpy ((char*)base + a * size_of_element,(char*)base + b * size_of_element,size_of_element);
memcpy ((char*)base + b * size_of_element, swap.zzz, size_of_element);
}

int main (void)
{
unsigned idx,array[] = {0,1,2,3,4,5,6,7,8,9};

swap_elements_of_array(array, sizeof array[0], 2, 5);

for (idx=0; idx < 10; idx++) {
    printf( "%u%c", array[idx], (idx==9) ? '\n' : ' ' );
    }
return 0;
}

The intention of the above fragment is to allow the highly optimised libc versions of memcpy (or inlining by the compiler) to take all the freedom they need. The alignment is crucial. If VLAs are not avalable (before C99) a macro can be composed, using a funky do-while.