Why is a struct with a deleted copy constructor not a POD type?

Question

We have the following method to test if our structure is a POD or not. It always returns true:

bool podTest() {
  struct podTest {
    int count;
    int x[];
  };

  return std::is_pod<podTest>::value;  //Always returns true
}

So far so good. Now we make one change and delete the copy constructor:

bool podTest() {
  struct podTest {
    podTest(const podTest&) = delete;
    int count;
    int x[];
  };

  return std::is_pod<podTest>::value;  //Always returns false
}

This always returns false. After reading over the definition of is_pod I am still struggling to understand what requirement it violates. What am I missing?

This is being compiled on godbolt using gcc 6.1, with -std=c++14

Answer 1

Aha!

From [class]:

A POD struct is a non-union class that is both a trivial class and a standard-layout class

where a trivial class is:

A trivial class is a class that is trivially copyable and has one or more default constructors (12.1), all of which are either trivial or deleted and at least one of which is not deleted.

But in [class.copy]:

If there is no user-declared constructor for class X, a non-explicit constructor having no parameters is implicitly declared as defaulted (8.4).

Your podTest, when you explicitly deleted the copy constructor, has no default constructor. So it's not a trivial class, so it's not a POD. If you add in:

podTest() = default;

Then it'll become a POD again.

Answer 2

Because deleted copy constructors are allowed for POD types only after C++14. I would assume, you are compiling your code in C++11 mode.