Derive class A from a template class Base<A> so that Base<A> can use A::B?

Question

template <typename T>
class Base
{
private:
    typename T::B c;
};

class A : public Base<A>
{
public:
    class B;
};

Is something like this even possible? VC++ 2013 says B is not a member of A.

Answer 1

THE STORY

As stated in the comments directly on your question what you are trying to accomplish simply isn't possible since it's illegal to refer to an incomplete type (which T = A is inside Base).

---

THE WORKAROUND

The common workaround in cases of CRTP is to use a trait to denote members that should be available in both Derived and Base, but which are not meant to be declared/defined in Base.

Even though this isn't really equivalent of what you are trying to accomplish it is very close to it, and follow somewhat equivalent semantics.

template<class> struct some_trait;       // primary template

template <class T>
struct A : some_trait<T> {               // (1)
  typename some_trait<T>::C a;
};

template<> struct some_trait<struct B> { // (2)
  class C { };
};

struct B : A<B> {                        // (3)
  C b;
};

int
main (int argc, char *argv[])
{
  B a; // legal
}

---

WHY DOES THE WORKAROUND... WORK?

To cut it short; we are no longer trying to access the internals of an incomplete type.

some_trait<B> is a complete type directly after it's definition (ie. the specialization) marked (2), and because of this it can be used by (1) and (3) without causing any issues.