sizeof std::aligned_storage and std::aligned_union - c++

Question

Given the following code:
#include <iostream>
#include <type_traits>
int main() {
std::aligned_storage<sizeof(double), alignof(double)> storage;
std::aligned_union<sizeof(double), double> union_storage;
std::cout << sizeof(storage) << '\n';
std::cout << sizeof(union_storage) << '\n';
std::cout << sizeof(double) << '\n';
}
I expect sizeof(storage) and sizeof(union_storage) to be greater or equal to sizeof(double) since they have to be able to hold a double. However, I get the output
1
1
8
clang-3.8 and gcc-5.3 both produce this output.
Why does sizeof return an incorrect size?
If I use placement new to put a double into storage or union_storage would that be undefined behavior?

Answer 1

std::aligned_storage and std::aligned_union are type traits which provide a member type which is the actual type of the storage.
Thus, placing a double in the memory of the actual trait type would indeed be UB because they're empty types with just a typedef member.
#include <iostream>
#include <type_traits>
int main()
{
using storage_type =
std::aligned_storage<sizeof(double), alignof(double)>::type;
using union_storage_type =
std::aligned_union<sizeof(double), double>::type;
storage_type storage;
union_storage_type union_storage;
std::cout << sizeof(storage_type) << '\n';
std::cout << sizeof(union_storage_type) << '\n';
std::cout << sizeof(storage) << '\n';
std::cout << sizeof(union_storage) << '\n';
std::cout << sizeof(double) << '\n';
return 0;
}
This gives:
8
8
8
8
8
Note: As #T.C. correctly noted: C++14 provides alias templates ending on _t for the std type traits (i.e. std::aligned_storage<L, A>::type === std::aligned_storage_t<L,A>). The benefit is
No typename in template dependant context.
Less typing. ;)