紧凑和简单的std :: tuple反演

Question

我是元编程的新手.我看过其他类似的问题,但没有一个问题符合我的要求.

这是我尝试反转std :: tuple.我的主要问题是反转输入元组中的数据.

反转指数的逻辑不合适,我无法从这个阶段开始.

到目前为止的代码:

//===========================================================!
// type inversion of a tuple

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;
};
//===========================================================!


template <typename First, typename ...Tails>
auto inverse(std::tuple<First, Tails...> & data) 
-> decltype(tuple_reverse<std::tuple<First,Tails...>>::type)
    {
        using reverse_tup = tuple_reverse<std::tuple<First, Tails...>>::type;
        static_assert(false, "Confused!")
        return reverse_tup();
    }

期待一个紧凑而简单的解决方案.

Answer 1

这是使用C++ 14的可能解决方案:

template <typename T, std::size_t... indices>
auto invert(T &&tuple, std::index_sequence<indices...>) {
  // Using decay_t as the argument must be a tuple, and this shortens the code
  using tuple_t = std::decay_t<T>;
  constexpr auto tuple_size = std::tuple_size<tuple_t>{};
  return std::tuple<std::tuple_element_t<tuple_size - indices - 1, tuple_t>...>(
      std::get<tuple_size - indices - 1>(std::forward<T>(tuple))...);
}

template <typename T>
auto invert(T &&tuple) {
  return invert(std::forward<T>(tuple),
                std::make_index_sequence<std::tuple_size<std::decay_t<T>>{}>());
}

演示.对于C++ 11,可以使用相同的过程,但make_index_list必须提供辅助模板.