Obtaining a type from a base class

Question

Suppose we have

class Derived : public Base1, public Base2, ..., public Base<Ts...>> {};

How to write a class that obtains the type std::tuple<Ts...> from Derived? Perhaps Base needs to be passed as a template argument too, to specify which base class to search for.

Providing my concrete example here will be overwhelming to read. Let's just say I need the Ts... in order to count how many of them satisfy a certain predicate. This number is the constexpr int I need for each derived class, but I need to obtain the Ts... first.

Answer 1

Since you publically and unambiguously inherit from a Base<Ts...>, you can use template argument deduction to get the Ts...: https://godbolt.org/z/oWfe3z16j

#include <type_traits>
#include <tuple>

namespace detail {
template<template<typename...> class Base, typename... Args> std::type_identity<std::tuple<Args...>> extract_base_args_impl(Base<Args...>*);
}

template<template<typename...> class Base, class Derived>
using extract_base_args_t = decltype(detail::extract_base_args_impl<Base>(static_cast<std::remove_cv_t<Derived>*>(nullptr)))::type;

template<template<typename...> class Base, class Derived>
struct extract_base_args {};

template<template<typename...> class Base, class Derived>
requires(requires { typename extract_base_args_t<Base, Derived>; })
struct extract_base_args<Base, Derived> {
    using type = extract_base_args_t<Base, Derived>;
};

Where you can make Base whatever template class you like.