Obtain using definition from templated class - c++

Question

Given the following classes:
// Some random class
class A { };
// A templated class with a using value in it.
template<class TYPE_B>
class B {
public:
using TYPE = TYPE_B;
};
Next we use these two classes in class C. But if we are using B as the template parameter we would like to obtain the TYPE defined in it.
template<class TYPE_C>
class C {
// A check to see if we have a class of type B
static constexpr bool IS_B = std::is_same<B<int32_t>, TYPE_C>::value ||
std::is_same<B<int64_t>, TYPE_C>::value;
public:
// This is what not works. How to get B::TYPE here?
using TYPE = std::conditional<IS_B, TYPE_C::TYPE, TYPE_C>;
};
Class C would we used like:
C<A> ca;
C<B<int32_t>> cb32;
C<B<int64_t>> cb64;
I am compiling this in GCC. My fear what I would like not have to do is to use the std::is_same statement for each type used with B. Put that in the std::conditional. Are there any alternatives?

Answer 1

You have two issues in your code:
You are missing a typename before TYPE_C::TYPE. Since TYPE_C::TYPE is a dependent name, you need to use typename to tell the compiler that you are looking for a type.
You cannot use TYPE_C::TYPE ins the std::conditional1 expression because when TYPE_C is not B<>, that expression is invalid, but will still be evaluated.
One simple solution is to use an intermediate template to access the type, using template specialization, e.g.
template <class T>
struct get_C_type {
using type = T;
};
template <class T>
struct get_C_type<B<T>> {
using type = T;
};
template<class TYPE_C>
class C {
public:
// you still need a typename here
using TYPE = typename get_C_type<TYPE_C>::type;
};
1 You probably want to use std::conditional_t or std::conditional<>::type here.