Question

Is this struct a POD in C++11?

struct B
{
  int a;
  B(int aa) : a(aa) {}
  B() = default;
};

Note that this question is explicit about C++11. I know that this class is not a POD in C++98 nor C++03.

For an explanation of POD in C++11, see trivial vs. standard layout vs. POD

(Inspired by this question: Is there a compile-time func/macro to determine if a C++0x struct is POD? )

Answer 1

Yes, it is a POD according to the new rules.

If you look up paragraph §8.4.2/4 of the new standard, you can see that if a constructor is defaulted on the first declaration, it is not user-provided:

Explicitly-defaulted functions and implicitly-declared functions are
collectively called defaulted functions, and the implementation shall
provide implicit definitions for them (§12.1 §12.4, §12.8), which might
mean defining them as deleted. A special member function is
user-provided if it is user-declared and not explicitly defaulted or
deleted on its first declaration. (…)

You can use the std::is_pod type trait to have the compiler test this for you with static_assert.

static_assert(std::is_pod<B>::value, "B should be a POD");