Memory alignment check

Question

I want to check whether an allocated memory is aligned or not. I am using _aligned_malloc(size, align); And it returns a pointer. Can I check it by simply dividing

Answer 1

On a modern Unix system a pointer returned by malloc is most likely 16 byte aligned as this is required for things like SSE. To check for alignment of a power of 2 you can use:

((unsigned long)p & (ALIGN - 1)) == 0

This is simply a faster version of (p % ALIGN) == 0. (If ALIGN is a constant your compiler will probably automatically use the faster version above.)

Answer 2

An "aligned" pointer by definition means that the numeric value of the pointer is evenly divisible by N (where N is the desired alignment). To check this, cast the pointer to an integer of suitable size, take the modulus N, and check whether the result is zero. In code:

bool is_aligned(void *p, int N)
{
    return (int)p % N == 0;
}

If you want to check the pointer value by hand, just look at the hex representation of the pointer and see whether it ends with the required number of 0 bits. A 16 byte aligned pointer value will always end in four zero bits, for example.

Answer 3

Memory returned by malloc is aligned for everything (ie, it generally uses the an alignment that works for everything)*. That means, if you have an alignment issue, it is something else.

http://www.delorie.com/gnu/docs/glibc/libc_31.html

http://msdn.microsoft.com/en-us/library/ms859665.aspx

(There appears to be exceptions for higher orders of alignment, which is an unusual requirement anyway.)