我该如何评价可变参数模板模板参数？

Question

可变参数模板模板参数接受任何模板:

template<typename T>
struct Test1 {
    using type = int;
};

template<typename T, typename T1>
struct Test2 {
    using type = char*;
};

template<template<typename...S> class BeCurry>
struct Currying {
};

using curry  = Currying<Test1>;
using curry2 = Currying<Test2>;

我想要Currying模板模板类.
这意味着:如果参数接受一个模板参数Test1,curry::apply<T>::type get Test1<T>::type.如果参数接受两个模板参数Test2,curry2::apply<T0>则为"部分"模板,curry2::apply<T0>::apply<T1>::type get Test2<T0,T1>::type

这有可能实现吗？因为我无法查询模板模板参数的内部参数num:

template<template<typename... S> class BeCurry>
struct Currying {
    enum { value = sizeof...(S) }; // error!
};

Answer 1

简单的解决方案是：

template
    <
        template <typename...> class BeCurry,
        typename... Params
    >
struct Currying
{
    template <typename... OtherParams>
    using curried = BeCurry<Params..., OtherParams...>;

    template <typename... OtherParams>
    using type = typename curried<OtherParams...>::type;

    template <typename... NewParams>
    using apply = Currying<curried, NewParams...>;
};

但由于编译错误（至少在gcc下），它不适用于Test1和等模板。此问题的解决方法如下所示：Test2

template
    <
        template <typename...> class BeCurry,
        typename... Params
    >
struct Curry
{
    using type = BeCurry<Params...>;
};

template
    <
        template <typename...> class BeCurry
    >
struct Curry<BeCurry>
{
    using type = BeCurry<>;
};

现在线路

template <typename... OtherParams>
using curried = BeCurry<Params..., OtherParams...>;

应该用线条代替

template <typename... OtherParams>
using curried = typename Curry<BeCurry, Params..., OtherParams...>::type;

使用示例：

#include <iostream>
#include <typeinfo>

template <typename T>
void print_type(T t)
{
    std::cout << typeid(t).name() << std::endl;
}

// ...

print_type(Currying<Test1>::type<int>{});
print_type(Currying<Test1>::apply<int>::type<>{});
print_type(Currying<Test2>::type<int, char>{});
print_type(Currying<Test2>::apply<int>::type<char>{});
print_type(Currying<Test2>::apply<int>::apply<char>::type<>{});
print_type(Currying<Test2>::apply<int, char>::type<>{});

完整示例位于ideone。