如何从可变参数包模板 类 中的每个类型恢复非类型名模板参数？

Question

我正在尝试从模板 classes 的可变包中恢复非类型名模板参数（其中每个 class 都有一个非类型参数），以便我可以将它们用作另一种类型的整数序列。

下面的代码显示了我所拥有的。整数序列/成员序列是从元组中抄袭的。

template<std::size_t... Integers>
struct size_t_sequence {
    using type = size_t_sequence<Integers...>;
};

template <std::size_t, typename>
struct push_size_t_sequence;

template <std::size_t I, std::size_t... Integers>
struct push_size_t_sequence<I, size_t_sequence<Integers...>>
    : size_t_sequence<I, Integers...> {};

template <std::size_t N, std::size_t Head, std::size_t... Tail>
struct make_size_t_sequence
    : push_size_t_sequence<Head, typename make_size_t_sequence<N - 1, Tail...>::type>::type {};

template<std::size_t I, std::size_t... OneLeft>
struct make_size_t_sequence <2, I, OneLeft...> :
    push_size_t_sequence<I, size_t_sequence<OneLeft...>>::type {};



template<typename... Members>
struct member_sequence {
    using type = member_sequence<Members...>;
};
template <typename, typename>
struct push_member_sequence;

template <typename M, typename... Members>
struct push_member_sequence<M, member_sequence<Members...>>
    : member_sequence<M, Members...> {};

template <std::size_t N, typename Head, typename... Tail>
struct make_member_sequence
    : push_member_sequence<Head, typename make_member_sequence<N - 1, Tail...>::type>::type {};

template<typename M, typename... OneLeft>
struct make_member_sequence <2, M, OneLeft...> :
    push_member_sequence<M, member_sequence<OneLeft...>>::type {};


template<typename>
struct unpack_sequence_impl;

template<template<std::size_t> class... T, std::size_t... DimensionSizes>
struct unpack_sequence_impl<member_sequence<T<DimensionSizes>...>> {
    using member_types = member_sequence<T<DimensionSizes>...>;
    using size_types = size_t_sequence<DimensionSizes...>;
};

template<typename... Members>
struct unpack_sequence : 
    unpack_sequence_impl<make_member_sequence<sizeof...(Members), Members...>> {
    using base_t = unpack_sequence_impl<make_member_sequence<sizeof...(Members), Members...>>;
    using member_types = typename base_t::member_types;
    using size_types = typename base_t::size_types;
};


template<std::size_t N>
class example {
    int s = N;
};

int main()
{
    auto mem_sequence = make_member_sequence<3, example<3>, example<2>, example<1>>::type();
    auto integer_sequence = make_size_t_sequence<3, 3,2,1>::type();
    auto un_mem_sequence = unpack_sequence<example<3>, example<2>, example<1>>::member_types();
    auto un_size_sequence = unpack_sequence<example<3>, example<2>, example<1>>::size_types();
}

mem_sequence和integer_sequence的类型分别是member_sequence<example<3>,example<2>,example<1>>和size_t_sequence<3,2,1>。 un_mem_sequence和un_size_sequence的类型应该是一样的。

我怎样才能做到这一点？

感谢您的帮助！

蒂姆

编辑：

为了澄清，我试图完成的是从一个模板 class 恢复模板参数以在另一个模板中使用它们。以下是三个模板 classes：MyObject、MyTuple、MyPack。 MyTuple 将 MyObject 个对象作为其模板参数。我想恢复 MyObject 模板参数以用作 MyPack 对象的模板参数。

template<int N>
MyObject;

template<int... Ns>
MyPack;

template<typename... MyObjects>
MyTuple {};



int main() {
    MyTuple<MyObject<1>,MyObject<2>,MyObject<3>> a;
    MyPack<1,2,3> b;
}

所以我想从 MyObject 的 MyTuple 中提取参数用于创建 MyPack。

编辑 2：

第二个澄清：MyTuple 不仅接受 MyObject 类型，还接受任何具有一个 int 模板参数的类型。

template<int N>
MyObject;

template<int N>
MyObject2;

template<int... Ns>
MyPack;

template<typename... MyObjects>
MyTuple {};



int main() {
    MyTuple<MyObject<1>,MyObject<2>,MyObject2<1>,MyObject2<2>> a;
    MyPack<1,2,1,2> b;
}

Answer 1

template <typename>
struct MakeMyPack;

template <int... Ns>
struct MakeMyPack<MyTuple<MyObject<Ns>...>>
{
    using type = MyPack<Ns...>;
};

DEMO

Answer 2

For clarification what I am trying to accomplish is recovering the template arguments from one template class to use them in another. Below are three template classes: MyObject, MyTuple, MyPack. MyTuple takes MyObject objects as its template parameters. I want to recover the MyObject template parameters to use as the template argument for a MyPack object.

只是为了好玩，我提出了一个更通用的解决方案，带有双可变参数模板（模板模板和值模板），不受 MyTuple 和 MyObject

的限制

template <typename>
struct getMyPack;

template <template <typename...> class C,
          template <int> class ... Cs, int ... Is>
struct getMyPack<C<Cs<Is>...>>
 { using type = MyPack<Is...>; };

下面是一个完整的编译示例

#include <type_traits>

template <int>
struct MyObject1
 { };

template <int>
struct MyObject2
 { };

template <int...>
struct MyPack
 { };

template <typename...>
struct MyTuple
 { };

template <typename>
struct getMyPack;

template <template <typename...> class C,
          template <int> class ... Cs, int ... Is>
struct getMyPack<C<Cs<Is>...>>
 { using type = MyPack<Is...>; };


int main ()
 {
   using T0 = MyTuple<MyObject1<1>, MyObject2<2>, MyObject1<3>>;
   using T1 = MyPack<1, 2, 3>;
   using T2 = typename getMyPack<T0>::type;

   static_assert( std::is_same<T1, T2>::value, "!" );
 }

-- 编辑--

Second clarification: MyTuple does not just take MyObject types but any type that has one template parameter of int.

正如我所怀疑的那样。

我的解决方案不适合 MyObject，所以应该可行。

修改前面的示例以显示它。