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?
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>
{};
If you want don't want your program to compile with foo<void> and don't mind ugly SFINAE, then here is an ugly solution with no specialization:
template< typename T, typename = std::enable_if_t<!std::is_same_v<void, T>>>
struct foo
{
using reference = T&;
using const_reference = T const&;
};