Using syntax to expose base class alias templates and variable templates in derived class?

Question

Consider a base template class and a derived template class in a heavy template metaprogramming context (simplified here for readability and to focus on the problem).

template <class T>
struct base {
    using type = T;
    static constexpr int value = 0;
    template <class... Args>
    constexpr void function(Args&&...) {}
    template <class U>
    using alias_template = base<U>;
    template <class U>
    static constexpr int variable_template = 0; 
};

template <class T>
struct derived: base<T> {
    using typename base<T>::type;           // Using base typedef
    using base<T>::value;                   // Using base static data member
    using base<T>::function;                // Using base function members (regardless of template or not)
    //using typename base<T>::alias_template; // DOES NOT SEEM TO WORK
    //using base<T>::variable_template;       // DOES NOT SEEM TO WORK

    using typedef_test = type;                                          // Working
    static constexpr int value_test = value;                            // Working
    using alias_template_test = alias_template<T>;                      // ERROR
    static constexpr int variable_template_test = variable_template<T>; // ERROR
};

QUESTION: Is there a using syntax to expose alias templates and variables templates inherited from a base class in order to make the currently erroneous lines compile? Is there any workaround to avoid to specify base<T>:: every single time in the derived class (here it remains simple, but in my actual code, specifying everytime rapidly becomes annoying)?

score 2 · Accepted Answer · answered May 17 '20 at 19:22

using doesn't work this way for dependent member templates:

A using-declaration also can't be used to introduce the name of a dependent member template as a template-name (the template disambiguator for dependent names is not permitted)

I'm not sure if this will satisfy your requirements, but in order to make the alias work like a template, you could declare new member templates in the derived class defined in terms of the base's member templates:

template<typename U>
using alias_template = typename base<T>::template alias_template<U>;

template<typename U>
static constexpr auto variable_template = base<T>::template variable_template<U>;

However, IMO specifying base<T>:: every time is not a problem, and is cleaner than introducing new templates. You could even make it shorter with something like using Base = my_long_base_class_name<T>;

Using syntax to expose base class alias templates and variable templates in derived class?

1 Answers1