Question

如何反转元组中的类型？例如，我希望reverse_tuple<std::tuple<int, char, bool>>::type为std::tuple<bool, char, int>。我尝试了以下但是没有用。我做错了什么？

#include <type_traits>
#include <tuple>

template <typename... Ts>
struct tuple_reverse;

template <typename T, typename... Ts>
struct tuple_reverse<std::tuple<T, Ts...>>
{
    using type = typename tuple_reverse<
                            std::tuple<
                               typename tuple_reverse<std::tuple<Ts..., T>>::type
                            >
                          >::type;
};

template <typename T>
struct tuple_reverse<std::tuple<T>>
{
    using type = std::tuple<T>;
};

int main()
{
    using result_type = std::tuple<int, bool, char>;
    static_assert(
        std::is_same<
            tuple_reverse<var>::type, std::tuple<char, bool, int>
        >::value, ""
    );
}

以下是我的错误：

<子>

prog.cpp: In instantiation of ‘struct tuple_reverse<std::tuple<char, int, bool> >’:
  prog.cpp:15:34: recursively required from ‘struct tuple_reverse<std::tuple<bool, char, int> >’
  prog.cpp:15:34: required from ‘struct tuple_reverse<std::tuple<int, bool, char> >’
  prog.cpp:29:31: required from here
  prog.cpp:15:34: error: no type named ‘type’ in ‘struct tuple_reverse<std::tuple<int, bool, char> >’
  prog.cpp: In function ‘int main()’:
  prog.cpp:30:9: error: template argument 1 is invalid

Answer 1

你做错了什么在这里：

using type = typename tuple_reverse<
                        std::tuple<
                           typename tuple_reverse<std::tuple<Ts..., T>>::type
                        >
                      >::type;

从内到外看，你重新排序元组元素：tuple<Ts..., T>，然后你试图反转它，然后你把结果放在tuple，然后你试图反转那 ......嗯？！：）

这意味着每次实例化tuple_reverse时，都会给它一个相同大小的元组，因此它永远不会完成，并且会永远地递归实例化它自己。（然后，如果该递归甚至完成，你将结果元组类型放入一个元组中，所以你有一个包含N元素元组的单元素元组，并反转它，这没有任何作用，因为反转单元素元组是一个无操作）。

你想剥离其中一个元素，然后反转剩下的元素，然后再将它连接起来：

using head = std::tuple<T>;
using tail = typename tuple_reverse<std::tuple<Ts...>>::type;

using type = decltype(std::tuple_cat(std::declval<tail>(), std::declval<head>()));

你不需要将它包装在一个元组中并再次反转它。）

你还应该处理空元组的情况，所以整个事情是：

template <typename... Ts>
struct tuple_reverse;

template <>
struct tuple_reverse<std::tuple<>>
{
    using type = std::tuple<>;
};

template <typename T, typename... Ts>
struct tuple_reverse<std::tuple<T, Ts...>>
{
  using head = std::tuple<T>;
  using tail = typename tuple_reverse<std::tuple<Ts...>>::type;

  using type = decltype(std::tuple_cat(std::declval<tail>(), std::declval<head>()));
};

我会这样做。

使用C ++ 14

获取类型

template<typename T, size_t... I>
struct tuple_reverse_impl<T, std::index_sequence<I...>>
{
  typedef std::tuple<typename std::tuple_element<sizeof...(I) - 1 - I, T>::type...> type;
};

// partial specialization for handling empty tuples:
template<typename T>
struct tuple_reverse_impl<T, std::index_sequence<>>
{
  typedef T type;
};

template<typename T>
struct tuple_reverse<T>
: tuple_reverse_impl<T, std::make_index_sequence<std::tuple_size<T>::value>>
{ };

或者您可以编写一个函数来反转实际的元组对象，然后使用decltype(reverse(t))来获取类型。在C ++ 14中反转类似元组的对象：

template<typename T, size_t... I>
auto
reverse_impl(T&& t, std::index_sequence<I...>)
{
  return std::make_tuple(std::get<sizeof...(I) - 1 - I>(std::forward<T>(t))...);
}

template<typename T>
auto
reverse(T&& t)
{
  return reverse_impl(std::forward<T>(t),
                      std::make_index_sequence<std::tuple_size<T>::value>());
}

在C ++ 11中使用<integer_seq.h>并添加返回类型并使用remove_reference从元组类型中删除引用（因为tuple_size和tuple_element不能使用对元组的引用）：

template<typename T, typename TT = typename std::remove_reference<T>::type, size_t... I>
auto
reverse_impl(T&& t, redi::index_sequence<I...>)
-> std::tuple<typename std::tuple_element<sizeof...(I) - 1 - I, TT>::type...>
{
    return std::make_tuple(std::get<sizeof...(I) - 1 - I>(std::forward<T>(t))...);
}

template<typename T, typename TT = typename std::remove_reference<T>::type>
auto
reverse(T&& t)
-> decltype(reverse_impl(std::forward<T>(t),
                        redi::make_index_sequence<std::tuple_size<TT>::value>()))
{
    return reverse_impl(std::forward<T>(t),
                        redi::make_index_sequence<std::tuple_size<TT>::value>());
}

Answer 2

未测试。

template < typename Tuple, typename T >
struct tuple_push;

template < typename T, typename ... Args >
struct tuple_push<std::tuple<Args...>, T>
{
    typedef std::tuple<Args...,T> type;
};

template < typename Tuple >
struct tuple_reverse;

template < typename T, typename ... Args >
struct tuple_reverse<std::tuple<T, Args...>>
{
    typedef typename tuple_push<typename tuple_reverse<std::tuple<Args...>>::type, T>::type type;
};

template < >
struct tuple_reverse<std::tuple<>>
{
    typedef std::tuple<> type;
};

无论如何，还有什么东西。

这也只会扭转类型，这似乎就是你所追求的。反转实际元组将涉及函数，而不是元函数。

Answer 3

我在处理任意类型的反转模板参数时遇到了这个问题。

Jonathan Wakely的回答对元组很有用，但万一其他人需要反转任何类型，即T<P1, P2, ..., Pn>到T<Pn, Pn-1, ..., P1>，这就是我提出的问题（{{3 }}）。

namespace Details
{
    /// Get the base case template type `T<>` of a templated type `T<...>`
    template<typename>
    struct templated_base_case;

    template <template<typename...> class T, typename... TArgs>
    struct templated_base_case<T<TArgs...>>
    {
        using type = T<>;
    };

    /// Inner reverse logic.
    ///
    /// Reverses the template parameters of a templated type `T` such
    /// that `T<A, B, C>` becomes `T<C, B, A>`.
    ///
    /// Note that this requires `T<>` to exist.
    template<
        typename T,
        typename = typename templated_base_case<T>::type>
    struct reverse_impl;

    template<
        template <typename...> class T,
        typename... TArgs>
    struct reverse_impl<
        typename templated_base_case<T<TArgs...>>::type,
        T<TArgs...>>
    {
        using type = T<TArgs...>;
    };

    template<
        template<typename...> class T,
        typename first,
        typename... rest,
        typename... done>
    struct reverse_impl<
        T<first, rest...>,
        T<done...>>
    {
        using type = typename reverse_impl <T<rest...>, T<first, done...>>::type;
    };

    /// Swap template parameters of two templated types.
    ///
    /// `L<A, B, C> and R<X, Y, Z>` become `L<X, Y, Z> and R<A, B, C>`.
    template<typename L, typename R>
    struct swap_template_parameters;

    template<
        template<typename...> class L,
        template<typename...> class R,
        typename... x,
        typename... y>
    struct swap_template_parameters<L<x...>, R<y...>>
    {
        using left_type = L<y...>;
        using right_type = R<x...>;
    };
}

/// Parameter pack list of types
template <typename... Args>
struct type_list { };

/// Reverses the arguments of a templates type `T`.
///
/// This uses a `type_list` to allow reversing types like std::pair
/// where `std::pair<>` and `std::pair<T>` are not valid.
template<typename T>
struct reverse_type;

template<template<typename...> class T, typename... TArgs>
struct reverse_type<T<TArgs...>>
{
    using type = typename Details::swap_template_parameters<
        T<TArgs...>,
        typename Details::reverse_impl<type_list<TArgs...>>::type>::left_type;
};

可以组合一些实现逻辑，但我试图在这里尽可能清楚。

reverse_type可以应用于元组：

using my_tuple = std::tuple<int, bool, char>;

static_assert(
    std::is_same<
        typename reverse_type<my_typle>::type,
        std::tuple<char, bool, int>>::value,
    "");

或其他类型：

/// Standard collections cannot be directly reversed easily
/// because they take default template parameters such as Allocator.
template<typename K, typename V>
struct simple_map : std::unordered_map<K, V> { };

static_assert(
    std::is_same<
        typename reverse_type<simple_map<std::string, int>>::type,
        simple_map<int, std::string>>::value,
    "");

Reversal logic taken from here

Answer 4

出于兴趣，您是否真的想要反转元组类型，或者只是以相反的顺序处理每个元素（在我的项目中更常见）？

#include <utility>
#include <tuple>
#include <iostream>

namespace detail {

    template<class F, class Tuple, std::size_t...Is>
    auto invoke_over_tuple(F &&f, Tuple &&tuple, std::index_sequence<Is...>) {
        using expand = int[];
        void(expand{0,
                    ((f(std::get<Is>(std::forward<Tuple>(tuple)))), 0)...});
    }


    template<class Sequence, std::size_t I>
    struct append;
    template<std::size_t I, std::size_t...Is>
    struct append<std::index_sequence<Is...>, I> {
        using result = std::index_sequence<Is..., I>;
    };

    template<class Sequence>
    struct reverse;

    template<>
    struct reverse<std::index_sequence<>> {
        using type = std::index_sequence<>;
    };

    template<std::size_t I, std::size_t...Is>
    struct reverse<std::index_sequence<I, Is...>> {
        using subset = typename reverse<std::index_sequence<Is...>>::type;
        using type = typename append<subset, I>::result;
    };
}

template<class Sequence>
using reverse = typename detail::reverse<Sequence>::type;

template
        <
                class Tuple,
                class F
        >
auto forward_over_tuple(F &&f, Tuple &&tuple) {
    using tuple_type = std::decay_t<Tuple>;
    constexpr auto size = std::tuple_size<tuple_type>::value;
    return detail::invoke_over_tuple(std::forward<F>(f),
                                     std::forward<Tuple>(tuple),
                                     std::make_index_sequence<size>());
};

template
        <
                class Tuple,
                class F
        >
auto reverse_over_tuple(F &&f, Tuple &&tuple) {
    using tuple_type = std::decay_t<Tuple>;
    constexpr auto size = std::tuple_size<tuple_type>::value;
    return detail::invoke_over_tuple(std::forward<F>(f),
                                     std::forward<Tuple>(tuple),
                                     reverse<std::make_index_sequence<size>>());
};

int main()
{
    auto t = std::make_tuple("1", 2, 3.3, 4.4, 5, 6, "7");
    forward_over_tuple([](auto &&x) { std::cout << x << " "; }, t);
    std::cout << std::endl;

    reverse_over_tuple([](auto &&x) { std::cout << x << " "; }, t);
    std::cout << std::endl;
}

如何反转元组类型中元素类型的顺序？

4 个答案: