Byte aligned malloc implementation

Usually when we call malloc we are returned with a address where the memory is allocated. But suppose if we want to ensure that the memory allocated is aligned with some value how can we implement such an malloc. Below is an implementation of that:

	/*
	* nMalloc.cpp
	*
	* Created on: Dec 5, 2014
	* Author: prasadnair
	*/
	#include <stdlib.h>
	#include <stdio.h>
	void * aligned_malloc(size_t size, int align) {
	/* alignment could not be less then zero */
	if (align < 0) {
	return NULL;
	}
	/* Allocate necessary memory area client request - size parameter -
	* plus area to store the address of the memory returned by standard
	* malloc().
	*/
	void *ptr;
	/* accomodating atleast one aligned memory area by adding align-1
	/ size of (void *) is added since we intent to keep the below address
	/ in p. This will be retrieved later during free
	*/
	void *p = malloc(size + align - 1 + sizeof(void*));
	if (p != NULL) {
	/* Address of the aligned memory according to the align parameter*/
	ptr = (void*) (((size_t)p + sizeof(void*) + align -1) & ~(align-1));
	/* store the address of the malloc() above the beginning of our total memory area.
	*/
	*((void**)((size_t)ptr - sizeof(void*))) = p;
	/* Return the address of aligned memory */
	return ptr;
	}
	return NULL;
	}
	void aligned_free(void *p) {
	/* Get the address of the memory, stored at the
	* start of our total memory area.
	*/
	void *ptr = *((void**)((size_t)p - sizeof(void*)));
	free(ptr);
	return;
	}
	int main ()
	{
	int align;
	for (align = 2; align < 5000000; align= align*2)
	{
	int *p = (int *) aligned_malloc (1024, align);
	// trying to see if its aligned properly by finding mode with align
	// mode operator should always return zero if memory is aligned to align
	printf (" %p %d \n", p, ((long int)p)%align);
	aligned_free (p);
	}
	return 0;
	}

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