Conditionally provide a using declaration

Question

Suppose I've got a class foo with template parameter T and I want to provide a using declaration for the reference and const-reference types corresponding to T:

template<typename T>
struct foo
{
    using reference = T&;
    using const_reference = T const&;
};

Is there a way to "enable" these using declerations only if T is not void without speclializing the whole class foo?

Answer 1

You could inherit from a base class with a specialization for void:

template<typename T>
struct typedefs {
    using reference = T&;
    using const_reference = T const&;
};

template<>
struct typedefs<void> {};

template<typename T>
struct foo : typedefs<T>
{};