Why is this not a POD?

Question

This is about std::is_pod, which detects whether a template is a plain old data type or not.

See the following code:

struct A { public:     int m1;     int m2; };  struct B { public:     int m1; private:     int m2; };  struct C { private:     int m1;     int m2; };  int main() {     std::cout << std::boolalpha;     std::cout << std::is_pod<A>::value << '\n'; // true     std::cout << std::is_pod<B>::value << '\n'; // false     std::cout << std::is_pod<C>::value << '\n'; // true } 

The 3 structs all look like POD to me. But apparently struct B is not. I don't understand why. To me, they all have a trivial constructor, move and copy operator. Destructor is certainly trivial too.

I blame it on using 2 access specifiers, but I can't find information about this.

Answer 1

According to the standard (9 Classes [class], emphasis mine):

A standard-layout class is a class that:

...

— has the same access control (Clause 11) for all non-static data members,

...

and

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

Your hunch is correct, because B.m1 and B.m2 are both non-static and have different access control.