Allocating aligned address and freeing them

 uintptr_t mask = ~(uintptr_t)(align - 1);
 void *mem = malloc(1024+align-1);
 void *ptr = (void *)(((uintptr_t)mem+align-1) & ~mask);
ptr is the aligned address.

Some time back I faced an interview ,There I was asked to write a custom malloc using malloc
and free program for a predefined aligned address.Here is the solution.The byte before the 
aligned byte is always empty.We will keep the offset of actual memory allocated by malloc.
But there we have to allocate total memory =(desired+alignment)  instead of  
(desired+alignment-1) downside is that it can store upto 2pow8 offsets.
 mallocX(size_t X,alignment Y)
{
 p= malloc(X+Y);
    ret = (p+Y) & ~(Y-1);
 *(ret-1) = ret - p;
    return ret;
}
similarly for free
freeX(memptr mem)
{
 free (mem - *(mem-1));
}