如何反转元组类型中元素类型的顺序

How do I reverse the order of element types in a tuple type?

本文关键字：类型 顺序 元素 元组 何反转      更新时间：2023-10-16

如何反转元组中的类型?例如，我想让reverse_tuple<std::tuple<int, char, bool>>::type变成std::tuple<bool, char, int>。我试着做以下，但它没有工作。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

你做错的地方在这里:

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>，然后你试着反转，然后你把结果放在tuple，然后你试着反转…啊? !:)

这意味着每次实例化tuple_reverse时，都给它一个相同大小的元组，因此它永远不会完成，并且永远递归地实例化自己。(然后，如果递归完成，您将结果元组类型放入元组中，这样您就有一个包含n元素元组的单元素元组，并将其反转，这没有任何作用，因为反转单元素元组是无操作的。)

您想要剥离其中一个元素，然后反转其余元素，并再次将其连接起来:

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>()));

你不需要把它包装在一个元组中，然后再反转:)

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>()));
};

我会用不同的方法。

获取类型，使用c++ 14

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一个实际的元组对象，然后使用decltype(reverse(t))来获取该类型。在c++ 14中反转一个类元组对象:

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>());
}

未经测试

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;
};

无论如何都有一些东西。

这也只是反转类型，这似乎是你想要的。反转一个实际的元组需要用到函数，而不是元函数。

我在为任意类型反转模板参数时遇到了这个问题。

Jonathan Wakely的答案对于元组非常有效，但如果其他人需要反转任何类型，即T<P1, P2, ..., Pn>到T<Pn, Pn-1, ..., P1>，这是我想到的(从这里取的反转逻辑)。

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;
};

一些实现逻辑可以组合在一起，但我在这里尽量使它尽可能清晰。

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,
    "");

更详细的说明

我结合了Faheem Mitha的SequenceHelper和Jonathan Wakely的tuple_reverse来反转元组类型和数据。此解决方案适用于c++11。

#include <tuple>
#include <algorithm>
#include <vector>
namespace TupleHelper {
namespace SequenceHelper {
// Tuple to parameter pack
// Original Author: Faheem Mitha https://stackoverflow.com/users/350713/faheem-mitha
// https://stackoverflow.com/questions/36612596/tuple-to-parameter-pack
//
// License: Creative Commons Attribution-ShareAlike (CC-BY-SA)
// https://www.ictrecht.nl/en/blog/what-is-the-license-status-of-stackoverflow-code-snippets
template<int ...>
struct seq { };
template<int N, int ...S>
struct gens : gens<N-1, N-1, S...> {};
template<int ...S>
struct gens<0, S...> {
    typedef seq<S...> type;
#if __GNUC__
    virtual ~gens() {} // this is only to avoid -Weffc++ warning
#endif
};
// end of Faheem Mitha's code
} // namespace SequenceHelper
// reverses the tuple types
// How do I reverse the order of element types in a tuple type?
// Original Author: Jonathan Wakely https://stackoverflow.com/users/981959/jonathan-wakely
// https://stackoverflow.com/a/17178399
//
// License: Creative Commons Attribution-ShareAlike (CC-BY-SA)
// https://www.ictrecht.nl/en/blog/what-is-the-license-status-of-stackoverflow-code-snippets
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>()));
};
// end of Jonathan Wakely's code
// reverses the content of the tuple too
namespace TupleReverseImpl {

// Helper functions to recursivly copy the source tuple to the target tuple
template <class ...Args>
void doCopyTypes( std::vector<void*>::size_type index,
                  std::vector<const void*> & content_source,
                  std::vector<void*> & content_target ) {}
template <class T, class ...Args>
void doCopyTypes( std::vector<void*>::size_type index,
                  std::vector<const void*> & content_source,
                  std::vector<void*> & content_target,
                  const T * t,
                  Args... args )
{
  // target and source vector do have the types in the same order
  // so we can cast here to the correct type, then copy it
  T * ptrSource = reinterpret_cast<T*>(const_cast<void*>(content_source.at(index)));
  T * ptrTarget = reinterpret_cast<T*>(content_target.at(index));
  *ptrTarget = *ptrSource;
  doCopyTypes( index+1, content_source, content_target, args... );
}
template <class Tuple, int ...S>
void copyTypes( std::vector<const void*> & content_source,
                std::vector<void*> & content_target,
                Tuple & tuple,
                SequenceHelper::seq<S...> )
{
  doCopyTypes( 0, content_source, content_target, &std::get<S>(tuple)... );
}
// Helper functions to fill a vector of pointers, to prepare copying
template <class V>
void dofillContent( V & content ) {}
template <class V, class T, class ...Args>
void dofillContent( V & content, T * t, Args... args )
{
  content.push_back( t );
  dofillContent( content, args... );
}
template <class V,class Tuple, int ...S>
void fill( V & content, Tuple & tuple, SequenceHelper::seq<S...> )
{
  dofillContent( content, &std::get<S>(tuple)... );
}
} // namespace TupleReverseImpl
/*
 * this reverses a tuple and its content
 *
 * std::tuple<int,bool,std::string> reverse = TupleHelper::reverseTuple( std::make_tuple( std::string( "one", true, 42 ) ) );
 */
template <class Tuple>
typename tuple_reverse<Tuple>::type reverseTuple( const Tuple & tuple )
{
    // declare return type
    typedef typename tuple_reverse<Tuple>::type REVERSE_TUPLE_TYPE;
    REVERSE_TUPLE_TYPE return_tuple;
    // create source and target pointer vectors for the copy action
    std::vector<const void*> contentSource;
    std::vector<void*> contentTarget;
    TupleReverseImpl::fill( contentSource, tuple, typename SequenceHelper::gens<std::tuple_size<Tuple>::value>::type() );
    TupleReverseImpl::fill( contentTarget, return_tuple, typename SequenceHelper::gens<std::tuple_size<REVERSE_TUPLE_TYPE>::value>::type() );
    // to be in the same order as contentTarget
    std::reverse(contentTarget.begin(), contentTarget.end() );

    // now copy everything
    TupleReverseImpl::copyTypes( contentSource, contentTarget, tuple, typename SequenceHelper::gens<std::tuple_size<Tuple>::value>::type() );
    return return_tuple;
}
} // namespace TupleHelper
int main()
{
    std::tuple<int,bool,std::string> reverse = TupleHelper::reverseTuple( std::make_tuple( std::string( "one", true, 42 ) ) );
}

出于兴趣，您是否真的想要反转元组类型，或者只是以反向顺序处理每个元素(就像我的项目中更常见的情况一样)?

#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;
}

#include <tuple>
template<typename T>
struct Reverse;
template<typename... Ts>
struct Reverse<std::tuple<Ts...>>
{
    using type = std::tuple<typename std::tuple_element<sizeof...(Ts) - 1 - i, std::tuple<Ts...>>::type...>;
};
// usage example
static_assert(std::is_same_v<typename Reverse<std::tuple<int, bool, double>>::type, std::tuple<double, bool, int>>);
using this method you can reverse the tuple type.
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