Answer by An̲̳̳drew for How to allocate aligned memory only using the standard library?

Here's an alternate approach to the 'round up' part. Not the most brilliantly coded solution but it gets the job done, and this type of syntax is a bit easier to remember (plus would work for alignment values that aren't a power of 2). The uintptr_t cast was necessary to appease the compiler; pointer arithmetic isn't very fond of division or multiplication.

void *mem = malloc(1024 + 15);
void *ptr = (void*) ((uintptr_t) mem + 15) / 16 * 16;
memset_16aligned(ptr, 0, 1024);
free(mem);