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，则curry2::apply<T0>是“部分”模板，curry2::apply<T0>::apply<T1>::type get Test2<T0,T1>::type

这可以实现吗？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Test2等模板。此问题的解决方法如下所示：

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

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

现在行

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

应替换为行

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<>{});

ideone的完整示例。

Answer 2

在some problems of my own之后，我想出了这个解决方案，它适用于任何模板类（也就是你在帖子中提供的模板类）。
该解决方案的核心是is_valid_specialization，用作判断currying过程是否完成的条件：

#include <iostream>
#include <type_traits>

template<template<typename...> class C, typename... T>
struct is_valid_specialization {
    typedef struct { char _; } yes;
    typedef struct { yes _[2]; } no;

    template<template<typename...> class D>
    static yes test(D<T...>*);
    template<template<typename...> class D>
    static no test(...);

    constexpr static bool value = (sizeof(test<C>(0)) == sizeof(yes));
};

namespace detail {

    template<template<typename...> class BeCurry, bool = false, typename... S>
    struct Currying {

        template<typename... T>
        using apply = Currying<BeCurry, is_valid_specialization<BeCurry, S..., T...>::value, S..., T...>;
    };

    template<template<typename...> class BeCurry, typename... S>
    struct Currying<BeCurry, true, S...> {

        template<typename... T>
        using apply = Currying<BeCurry, is_valid_specialization<BeCurry, S..., T...>::value, S..., T...>;

        using type = typename BeCurry<S...>::type;
    };
}

template<template<typename...> class BeCurry>
using Currying = detail::Currying<BeCurry, is_valid_specialization<BeCurry>::value>;

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

template<typename T1, typename T2>
struct Test2 { using type = char*; };

template<typename...>
struct Test3 { using type = double; };

using curry  = Currying<Test1>;
using curry2 = Currying<Test2>;
using curry3 = Currying<Test3>;

template<typename T>
void pretty_print(T) {
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

int main() {
    pretty_print(typename curry::apply<char>::type{});
    pretty_print(typename curry2::apply<int>::apply<char>::type{});
    pretty_print(typename curry3::type{});
}

ideone

上的输出

我该如何评价可变参数模板模板参数？

2 个答案: