Inheritance with variadic templates

Question

I'm trying to use inheritance with variadic templates. First, consider the following snippet as a skeleton I'd like to build upon.

#include <type_traits>
#include <iostream>

template <typename K> class A1;   // fwd decl

template <typename K,
        template <typename> class NN = A1,
        class = typename std::enable_if< std::is_base_of< A1<K>, NN<K> >::value >::type >
class BB;

template <typename K>
class A1 { public: friend class BB<K>; };

template <typename K>
class A2 : public A1<K> {
};

template <typename K, template <typename> class NN>
class BB<K,NN> {
    NN<K>* ref;
public:
    BB() : ref{ new NN<K>{} } { std::cout << "ctor...\n"; };
};

int main() {
    BB<size_t> b1{};    //use default A1
    BB<size_t, A2> b2{};
}

What we have here is that I can use the class BB with any class inherited from A1 that has exactly 1 template parameter.

Question: I'd like to modify this pattern in such a way that the template parameter NN can be any class inherited from A1<K> that may also have an arbitrary number of additional template parameters, how can I do it? The following snippet is the rough idea

template <typename K> class A1;   // fwd decl

template <typename K,
        template <typename, typename...> class NN = A1,
        class = typename std::enable_if< std::is_base_of< A1<K>, NN<K,typename...> >::value >::type,
        typename...Types >
class BB;

template <typename K>
class A1 { public: friend class BB<K>; };

template <typename K, typename V>
class A2 : public A1<K> {
    V local;
};

template <typename K,
        template <typename, typename...> class NN,
        typename...Types>
class BB<K,NN,void,Types...> {
    NN<K,Types...>* ref;
public:
    BB() : ref{ new NN<K,Types...>{} } { std::cout << "ctor...\n"; };
};

int main() {
    BB<size_t> b1{};    //use default A1
    BB<size_t, A2, char> b2{};  // can I specialize this way??
}

Is it possible to do? Any help is greatly appreciated. VS2017 or GCC is indifferent.

NN isn't a class, it is a template class. These are as different as a class and an variable; I don't know if your use of NN as a class betrays a misunderstanding of terminology or a misunderstanding of your problem. Second, your first code block doesn't look like it should work as it should require a ,void, but somehow it compiles. I'm puzzled. — Yakk - Adam Nevraumont
– Yakk - Adam Nevraumont, Commented May 22, 2018 at 17:15
yeah template class ofc. Where exactly did you expect a void? — Frank Buddha
– Frank Buddha, Commented May 22, 2018 at 18:48
BB<K,NN> I'd expect BB<K,NN,void>. I guess that you don't have to repeat defaulted template parameters in specializations or something? — Yakk - Adam Nevraumont
– Yakk - Adam Nevraumont, Commented May 22, 2018 at 19:10
Yeah that's the reason. In particular because it's the last one. — Frank Buddha
– Frank Buddha, Commented May 22, 2018 at 19:25

Yakk - Adam Nevraumont · Accepted Answer · 2018-05-22 17:27:27Z

template <typename K> class A1;   // fwd decl

template<class...>struct types_t {};
template <class K0, class Types=types_t<>, template<class...>class NN=A1, class=void>
class BB;

template <typename K>
class A1 { public: friend class BB<K>; };

template <class K, class other>
class A2 : public A1<K> {
};

template<class K, class...Ks>
class A3 : public A1<K> {};

template<class K0, class...Ks, template<class...>class NN>
class BB<
    K0,
    types_t<Ks...>,
    NN,
    std::enable_if_t<
        std::is_base_of<
            A1<K0>,
            NN<K0, Ks...>
        >{}
    >
>{
    NN<K0, Ks...>* ref = nullptr;
public:
    BB() : ref{ new NN<K0, Ks...>{} } { std::cout << "ctor...\n"; };
};

int main() {
    BB<size_t> b1{};    //use default A1
    BB<size_t, types_t<char>, A2> b2{};
    BB<size_t, types_t<int, char>, A3> b3{};
}

this compiles.

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