c++ - 在 C++ 中实现类型列表

Question

我正在使用模板研究 C++ 中的元编程，我正在尝试实现一个类型列表，并对其进行操作。

我将类型列表定义为可变参数类模板，并将操作定义为具有部分特化的模板结构。 Front、PopFront 和 PushFront 等操作工作正常，但是当我实例化 Back 和 Element(以索引类型列表以获取第 n 个元素)时，编译器提示我使用的类型不完整:

/**** typelist ****/
template <typename... Types>
struct Typelist
{
};

/**** get first element ****/
template <typename List>
struct Front;

template <typename Head, typename... Tail>
struct Front<Typelist<Head,Tail...>>
{
    typedef Head type;
};

template <typename List>
using FrontT = typename Front<List>::type;

/**** pop first element ***/
template <typename List>
struct PopFront;

template <typename Head, typename... Tail>
struct PopFront<Typelist<Head,Tail...>>
{
    using type = Typelist<Tail...>;
};

template <typename List>
using PopFrontT = typename PopFront<List>::type;

/**** push first element ****/
template <typename List, typename Element>
struct PushFront;

template <typename... Elements, typename Element>
struct PushFront<Typelist<Elements...>,Element>
{
    using type = Typelist<Element,Elements...>;
};

template <typename List, typename Element>
using PushFrontT = typename PushFront<List,Element>::type;

/**** get last element ****/ 
template <typename List>
struct Back;

template <typename... Head, typename Tail>
struct Back<Typelist<Head...,Tail>>
{
    typedef Tail type;
};

/**** indexing ****/
template <typename List, unsigned Index>           // recursive case
struct Element
{
    using type = typename Element<typename PopFront<List>::type, Index - 1>::type;
};

template <typename List>
struct Element<List,0>                             // base case
{
    typedef typename Front<List>::type type;
};

// template <typename... Types>                        // base case
// struct Element<Typelist<Types...>,0>
// {
//     using type = typename Front<Typelist<Types...>>::type;
// };

// template <typename... Types, unsigned Index>        // recursive case
// struct Element<Typelist<Types...>,Index>
// {
//     using type = typename Element<typename PopFront<Typelist<Types...>>::type, Index - 1>::type;
// };

template <typename List, unsigned Index>
struct ElementI : ElementI<PopFrontT<List>,Index - 1>
{
};

template <typename List>
struct ElementI<List,0> : Front<List>
{
};

template <typename List, unsigned Index>
using ElementT = typename Element<List,Index>::type;

我知道只要可以推断出参数，我就可以将模板参数包用于偏特化中的任何参数，所以我认为声明是正确的，对吧？

编辑
Element 现在可以工作，我在调用它时犯了一个拼写错误，Back 仍然没有，我不明白为什么。

编辑
这是测试类型列表和编译器(GCC 7.2)错误的代码(我稍微改变了类型列表的实现):

编辑编辑 编译器是 GCC 7.2

#include "typelist.hpp"
#include <type_traits>

int main(int argc, char **argv)
{
    Typelist<int, double, bool> tl;

    static_assert(std::is_same<typename Front<decltype(tl)>::type,int>::value, "not same");
    static_assert(std::is_same<typename PopFront<decltype(tl)>::type,Typelist<double,bool>>::value, "not same");
    static_assert(std::is_same<typename PushFront<decltype(tl),float>::type,Typelist<float,int,double,bool>>::value, "not same");
    /* compiler error */ static_assert(std::is_same<typename Back<decltype(tl)>::type,bool>::value, "not same");
    static_assert(std::is_same<typename ElementI<decltype(tl),0>::type,int>::value, "not same");
    static_assert(std::is_same<typename ElementI<decltype(tl),1>::type,double>::value, "not same");
    static_assert(std::is_same<ElementT<decltype(tl),2>,bool>::value, "not same");

    return 0;
}


main.cpp: In function 'int main(int, char**)':
main.cpp:11:61: error: invalid use of incomplete type 'struct Back<Typelist<int, double, bool> >'
     static_assert(std::is_same<typename Back<decltype(tl)>::type,bool>::value, "not same");
                                                             ^~~~
In file included from main.cpp:1:0:
typelist.hpp:48:8: note: declaration of 'struct Back<Typelist<int, double, bool> >'
 struct Back;
        ^~~~
main.cpp:11:70: error: template argument 1 is invalid
     static_assert(std::is_same<typename Back<decltype(tl)>::type,bool>::value, "not same");

Answer 1

与 clang++ (Apple clang 版本 11.0.3 (clang-1103.0.32.59))我收到以下消息:

t.cpp:51:8: error: class template partial specialization contains template parameters that cannot be deduced; this partial specialization will never be used [-Wunusable-partial-specialization]
struct Back<Typelist<Head...,Tail>>
       ^~~~~~~~~~~~~~~~~~~~~~~~~~~~
t.cpp:50:23: note: non-deducible template parameter 'Head'
template <typename... Head, typename Tail>
                      ^
t.cpp:50:38: note: non-deducible template parameter 'Tail'
template <typename... Head, typename Tail>
                                     ^

发件人: https://en.cppreference.com/w/cpp/language/parameter_pack (对不起，我没有主裁判)

In a primary class template, the template parameter pack must be the final parameter in the template parameter list.
In a function template, the template parameter pack may appear earlier in the list provided that all following parameters can be deduced from the function arguments, or have default arguments:

template<typename... Ts, typename U> struct Invalid; // Error: Ts.. not at the end

template<typename ...Ts, typename U, typename=void>
void valid(U, Ts...);     // OK: can deduce U
// void valid(Ts..., U);  // Can't be used: Ts... is a non-deduced context in this position

valid(1.0, 1, 2, 3);      // OK: deduces U as double, Ts as {int,int,int}

但是我们可以通过定义 Back 来解决这个问题。根据 Element .

/**** get last element ****/
template <typename List>
struct Back;

template <typename... Args>
struct Back<Typelist<Args...>>
{
    using type = typename Element<Typelist<Args...>, sizeof...(Args) - 1>::type;
};

问题:

为什么你有 template using除了 Back 之外的所有类型的语句和 Element ?