c++ - 在 C++ 中实现类型列表
问题描述
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专业化的模板结构。Front、PopFront 和 PushFront 之类的操作工作正常,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,编译器抱怨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?
EDIT Element 现在可以工作了,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");
解决方案
使用clang++
(Apple clang version 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>
^
来自:https ://en.cppreference.com/w/cpp/language/parameter_pack (对不起,我没有主要参考)
在主类模板中,模板参数包必须是模板参数列表中的最后一个参数。
在函数模板中,模板参数包可能出现在列表的前面,前提是所有以下参数都可以从函数参数中推导出来,或者具有默认参数:
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}
Back
但是我们可以通过定义来解决这个问题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:
为什么除了andtemplate using
之外的所有类型都有语句?Back
Element