展开std :: tuple以用作类的初始化程序

Question

我想使用a的所有元素std::tuple作为类的初始化器.有没有比std::get<i-th element>(std::tuple)为元组的每个元素做更简单的方法？

最低工作示例std::get:

#include <string>
#include <tuple>
#include <cassert>

struct A
{
    std::string string1;
    int intVal;
    std::string string2;
};

int main()
{
  std::tuple< std::string, int, std::string > myTuple("S1", 42, "S2");

  A myA{ std::get<0>(myTuple), std::get<1>(myTuple), std::get<2>(myTuple) };

  assert( myA.string1 == "S1" );
  assert( myA.intVal  == 42   );
  assert( myA.string2 == "S2" );
}

有关实例,请参见http://coliru.stacked-crooked.com/a/4a5d45dbf1461407

Answer 1

正如Kerrek SB评论的那样,已经提出了这个P0209R0的建议.因此,除非符合标准,否则您可以按照以下方式执行操作:

template<typename C, typename T, std::size_t... I>
decltype(auto) make_from_tuple_impl(T &&t, std::index_sequence<I...>) {
  return C{std::get<I>(std::forward<T>(t))...};
}

template<typename C, typename... Args, typename Indices = std::make_index_sequence<sizeof...(Args)>>
decltype(auto) make_from_tuple(std::tuple<Args...> const &t) {
  return make_from_tuple_impl<C>(t, Indices());
}

并将您的课程初始化为:

A myA{make_from_tuple<A>(myTuple)};

现场演示

您也可以手工制作,index_sequence并make_index_sequence在此处按照Jarod42的 建议在C++ 11中工作,并更改为:

namespace idx {
  template <std::size_t...> struct index_sequence {};

  template <std::size_t N, std::size_t... Is>
  struct make_index_sequence : make_index_sequence<N - 1, N - 1, Is...> {};

  template <std::size_t... Is>
  struct make_index_sequence<0u, Is...> : index_sequence<Is...> { using type = index_sequence<Is...>; };
}

template<typename C, typename T, std::size_t... I>
C make_from_tuple_impl(T &&t, idx::index_sequence<I...>) {
  return C{std::get<I>(std::forward<T>(t))...};
}

template<typename C, typename... Args, typename Indices = idx::make_index_sequence<sizeof...(Args)>>
C make_from_tuple(std::tuple<Args...> const &t) {
  return make_from_tuple_impl<C>(t, Indices());
}

现场演示