Memory alignment : how to use alignof / alignas?

Alignment is a restriction on which memory positions a value’s first byte can be stored. (It is needed to improve performance on processors and to permit use of certain instructions that works only on data with particular alignment, for example SSE need to be aligned to 16 bytes, while AVX to 32 bytes.)

Alignment of 16 means that memory addresses that are a multiple of 16 are the only valid addresses.

alignas

force alignment to the required number of bytes. You can only align to powers of 2: 1, 2, 4, 8, 16, 32, 64, 128, …

#include <cstdlib>
#include <iostream>

int main() {
    alignas(16) int a[4];
    alignas(1024) int b[4];
    printf("%p\n", a);
    printf("%p", b);
}

example output:

0xbfa493e0
0xbfa49000  // note how many more "zeros" now.
// binary equivalent
1011 1111 1010 0100 1001 0011 1110 0000
1011 1111 1010 0100 1001 0000 0000 0000 // every zero is just a extra power of 2

the other keyword

alignof

is very convenient, you cannot do something like

int a[4];
assert(a % 16 == 0); // check if alignment is to 16 bytes: WRONG compiler error

but you can do

assert(alignof(a) == 16);
assert(alignof(b) == 1024);

note that in reality this is more strict than a simple “%” (modulus) operation. In fact we know that something aligned to 1024 bytes is necessarily aligned to 1, 2, 4, 8 bytes but

 assert(alignof(b) == 32); // fail.

So to be more precise, “alignof” returns the greatest power of 2 to wich something is aligned.

Also alignof is a nice way to know in advance minimum alignment requirement for basic datatypes (it will probably return 1 for chars, 4 for float etc.).

Still legal:

alignas(alignof(float)) float SqDistance;

Something with an alignment of 16 then will be placed on the next available address that is a multiple of 16 (there may be a implicit padding from last used address).

Leave a Comment