Expose the most direct base class template name

Question

Consider the following code:

template <class...> struct base {};
template <class... T> struct intermediate: base<void, T...> {};
template <class... T> struct derived: base<T...>, intermediate<T...> {};

using type1 = derived<int>::intermediate::base; // works
using type2 = derived<int>::base; // ambiguous

Would there be a way to make it work without ambiguity so that derived<int>::base means the most "direct" base class in the hierarchy (in this example base<int>). Template metaprogramming is welcomed.

Could you change to `struct derived: wrapped>, intermediate {};`? — Jarod42, Feb 18 '20 at 18:02
Yes. I can change the hierachy as long as I can type `derived::base` and `derived::intermediate::base` — Vincent, Feb 18 '20 at 18:26
It would do `derived::wrapped::base`, `derived::intermediate::base`... :/ — Jarod42, Feb 18 '20 at 18:29

Jarod42 · Accepted Answer · 2020-02-18T17:57:26.163

1

If appropriate, you might add alias in derived, so you can use them from outside:

template <class... Ts> struct derived: base<Ts...>, intermediate<Ts...>
{
    using DirectBase = base<Ts...>;
    using IndirectBase = typename intermediate<Ts...>::base;
};

And then

using type1 = derived<int>::IndirectBase ;
using type2 = derived<int>::DirectBase ;

I see no ways to disambiguate.

There was proposal for std::bases std::direct_bases which might help, but rejected.

edited Feb 18 '20 at 17:57

answered Feb 18 '20 at 17:53

Jarod42

203,559
14
181
302

Unfortunately, I would like to keep the same name (in my real case, base might be a template template, and I would like to keep the same same, whatever it is) – Vincent Feb 18 '20 at 17:56
You cannot do `using base = base;` neither :-/ – Jarod42 Feb 18 '20 at 17:59
Normally indirect base is a base class of base class, not a "co-parent" – Swift - Friday Pie Feb 18 '20 at 18:07

Expose the most direct base class template name

1 Answers1