How to align pointer

Question

How do I align a pointer to a 16 byte boundary?

I found this code, not sure if its correct

char* p= malloc(1024);

if ((((unsigned long) p) % 16) !=          


        

Answer 1

Try this:

It returns aligned memory and frees the memory, with virtually no extra memory management overhead.

#include 
#include 

size_t roundUp(size_t a, size_t b) { return (1 + (a - 1) / b) * b; }

// we assume here that size_t and void* can be converted to each other
void *malloc_aligned(size_t size, size_t align = sizeof(void*))
{
    assert(align % sizeof(size_t) == 0);
    assert(sizeof(void*) == sizeof(size_t)); // not sure if needed, but whatever

    void *p = malloc(size + 2 * align);  // allocate with enough room to store the size
    if (p != NULL)
    {
        size_t base = (size_t)p;
        p = (char*)roundUp(base, align) + align;  // align & make room for storing the size
        ((size_t*)p)[-1] = (size_t)p - base;      // store the size before the block
    }
    return p;
}

void free_aligned(void *p) { free(p != NULL ? (char*)p - ((size_t*)p)[-1] : p); }

Warning:

I'm pretty sure I'm stepping on parts of the C standard here, but who cares. :P