BlackMamba BlackMamba - 11 months ago 66
C++ Question

Does alignas affect the value of sizeof?

#include <iostream>
using namespace std;

int main()
alignas(double) unsigned char c[1024]; // array of characters, suitably aligned for doubles
alignas(16) char d[100]; // align on 16 byte boundary
constexpr int n = alignof(int); // ints are aligned on n byte boundarie

Here is the code, for
alignas(double) unsigned char c[1024];
, it means the
should be aligned by
, the
So I think
should be
bytes, but the console output is

So I am confused. Who can tell me the reason?


The alignas keyword can be used to dictate alignment requirements. alignas(double) for example forces the variable to have the same alignment requirements as a double. On my platform, this will mean that the variable is aligned on 8 byte boundaries.

In your example, the whole array will get the alignment requirements so it's being aligned on 8 byte boundaries but this won't affect its size.

It is however possible that alignas changes the size of a composite data type when upholding the alignment requirements requires additional padding. Here's an example:

#include <iostream>
#include <cstddef>

struct Test
    char a;
    alignas(double) char b;

int main(int argc, char* argv[])
    Test test;
    std::cout << "Size of Struct: " << sizeof(Test) << std::endl;
    std::cout << "Size of 'a': " << sizeof(test.a) << std::endl;
    std::cout << "Size of 'b': " << sizeof(test.b) << std::endl;
    std::cout << "Offset of 'a': " << (int)offsetof(struct Test, a) << std::endl;
    std::cout << "Offset of 'b': " << (int)offsetof(struct Test, b) << std::endl;
    return 0;


Size of Struct: 16
Size of 'a': 1
Size of 'b': 1
Offset of 'a': 0
Offset of 'b': 8

The size of this structure is 16 bytes on my platform even though both members are just 1 byte in size each. So b didn't become bigger because of the alignment requirement but there is additional padding after a. You can see this by looking at the size and offset of the individual members. a is just 1 byte in size but b, due to our alignment requirements, starts after a 8 byte offset.

And the size of a struct must be a multiple of its alignment, otherwise arrays don't work. So if you set an alignment requirement that's bigger than the whole struct was to begin with (for example a struct containing only a single short and you apply alignas(double) to that data member), padding must be added after it.