What's the default value for a std::atomic?

Question

I find that in practice, with a variety of C++11/C++14 compilers, a std::atomic has an undefined initial value just as it would if it were a "raw" type. That is, we expect that for the expression

int a;

a may have any value. It also turns out to be true that for the expression

std::atomic< int > b;

b may also have any value. To say it another way,

std::atomic< int > b;         // b is undefined

is not equivalent to

std::atomic< int > b{ 0 };    // b == 0

or to

std::atomic< int > b{};       // b == 0

because in the latter two cases b is initialized to a known value.

My question is simple: where in the C++11 or C++14 spec is this behavior documented?

Answer 1

[atomics.types.generic]/5 says this about integral specializations:

The atomic integral specializations and the specialization atomic shall have standard layout. They shall each have a trivial default constructor and a trivial destructor. They shall each support aggregate initialization syntax.

Moreover, the primary template synopsis at the beginning of the same section normatively specifies the default constructor as:

atomic() noexcept = default;

The effects are defined in [atomic.types.operations]/4 as:

Effects: leaves the atomic object in an uninitialized state.