Can one use template specialization on a templated typedef?

Question

I would like to do something like the following (in c++11, c++14; not c++17):

template <class T>
using partner = void;

template<>
using partner<A> = X;

template<>
using partner<B> = Y;

template<>
using partner<C> = Z;

But I get a compilation error---

error: expected unqualified-id before ‘using’

---on the first template specialization.

Is such a thing possible? (I already know I can use a templated class with a using statement inside it. I'm hoping directly use the using statement without the class wrapper, since it's simpler and more elegant. If there's another simple, elegant solution, please share!)

Answer 1

You can't specialize alias templates.

You'll have to resort to normal, boring class template specialization:

template <class T> struct partner_t { using type = void; };
template <> struct partner_t<A> { using type = X; };
template <> struct partner_t<B> { using type = Y; };
template <> struct partner_t<C> { using type = Z; };

template <class T>
using partner = typename partner_t<T>::type;

---

Or we could get fancier

template <class T> struct tag { using type = T; };

template <class T> auto partner_impl(tag<T> ) -> tag<void>;
auto partner_impl(tag<A> ) -> tag<X>;
auto partner_impl(tag<B> ) -> tag<Y>;
auto partner_impl(tag<C> ) -> tag<Z>;

template <class T>
using partner = typename decltype(partner_impl(tag<T>{}))::type;