将std::tuple转换为std::array C++11

If I have std::tuple<double, double, double> (where the type is homogeneous), is there a stock function or constructor to convert to std::array<double>?

如果我有std::tuple (类型是齐次的)，是否有一个stock函数或构造函数可以转换为std::array ? ,>

Edit:: I was able to get it working with recursive template code (my draft answer posted below). Is this the best way to handle this? It seems like there would be a stock function for this... Or if you have improvements to my answer, I'd appreciate it. I'll leave the question unanswered (after all, I want a good way, not just a workable way), and would prefer to select someone else's [hopefully better] answer.

编辑:::我可以让它使用递归模板代码(我的草稿答案贴在下面)。这是最好的处理方法吗?这似乎有一个股票函数。或者如果你对我的答案有改进，我会很感激的。我将不回答这个问题(毕竟，我想要一种好的方法，而不仅仅是一种可行的方法)，我宁愿选择别人的(希望是更好的)答案。

Thanks for your advice.

谢谢你的建议。

5 个解决方案

#1

Converting a tuple to an array without making use of recursion, including use of perfect-forwarding (useful for move-only types):

将一个元组转换为数组而不使用递归，包括使用完全转发(仅对移动类型有用):

#include <iostream>
#include <tuple>
#include <array>

template<int... Indices>
struct indices {
    using next = indices<Indices..., sizeof...(Indices)>;
};

template<int Size>
struct build_indices {
    using type = typename build_indices<Size - 1>::type::next;
};

template<>
struct build_indices<0> {
    using type = indices<>;
};

template<typename T>
using Bare = typename std::remove_cv<typename std::remove_reference<T>::type>::type;

template<typename Tuple>
constexpr
typename build_indices<std::tuple_size<Bare<Tuple>>::value>::type
make_indices()
{ return {}; }

template<typename Tuple, int... Indices>
std::array<
  typename std::tuple_element<0, Bare<Tuple>>::type,
    std::tuple_size<Bare<Tuple>>::value
>
to_array(Tuple&& tuple, indices<Indices...>)
{
    using std::get;
    return {{ get<Indices>(std::forward<Tuple>(tuple))... }};
}

template<typename Tuple>
auto to_array(Tuple&& tuple)
-> decltype( to_array(std::declval<Tuple>(), make_indices<Tuple>()) )
{
    return to_array(std::forward<Tuple>(tuple), make_indices<Tuple>());
}

int main() {
  std::tuple<double, double, double> tup(1.5, 2.5, 4.5);
  auto arr = to_array(tup);
  for (double x : arr)
    std::cout << x << " ";
  std::cout << std::endl;
  return 0;
}

#2

You can do it non-recursively:

你可以不递归地做:

#include <array>
#include <tuple>
#include <redi/index_tuple.h>  // see below

template<typename T, typename... U>
  using Array = std::array<T, 1+sizeof...(U)>;

template<typename T, typename... U, unsigned... I>
  inline Array<T, U...>
  tuple_to_array2(const std::tuple<T, U...>& t, redi::index_tuple<I...>)
  {
    return Array<T, U...>{ std::get<I>(t)... };
  }

template<typename T, typename... U>
  inline Array<T, U...>
  tuple_to_array(const std::tuple<T, U...>& t)
  {
    using IndexTuple = typename redi::make_index_tuple<1+sizeof...(U)>::type;
    return tuple_to_array2(t, IndexTuple());
  }

See https://gitlab.com/redistd/redistd/blob/master/include/redi/index_tuple.h for my implementation of index_tuple, something like that is useful for working with tuples and similar variadic templates. A similar utility was standardised as std::index_sequence in C++14 (see index_seq.h for a standalone C++11 implementation).

关于我的index_tuple的实现，请参见https://git.com/redistd/redistd/blob/master/include/redi/index_tuple.h，类似的内容对于处理元组和类似的可变单元模板非常有用。一个类似的实用程序被标准化为std:: c++ 14中的index_sequence(参见index_seq)。h用于独立的c++ 11实现)。

#3

I would return the array instead of populating it by reference, so that auto can be used to make the callsite cleaner:

我将返回数组而不是通过引用填充它，这样auto就可以用来使callsite变得更干净:

template<typename First, typename... Rem>
std::array<First, 1+sizeof...(Rem)>
fill_array_from_tuple(const std::tuple<First, Rem...>& t) {
  std::array<First, 1+sizeof...(Rem)> arr;
  ArrayFiller<First, decltype(t), 1+sizeof...(Rem)>::fill_array_from_tuple(t, arr);
  return arr;
}

// ...

std::tuple<double, double, double> tup(0.1, 0.2, 0.3);
auto arr = fill_array_from_tuple(tup);

Realistically, NRVO will eliminate most performance concerns.

实际上，NRVO将消除大多数性能问题。

#4

#include <iostream>
#include <tuple>
#include <array>

template<class First, class Tuple, std::size_t N, std::size_t K = N>
struct ArrayFiller {
  static void fill_array_from_tuple(const Tuple& t, std::array<First, N> & arr) {
    ArrayFiller<First, Tuple, N, K-1>::fill_array_from_tuple(t, arr);
    arr[K-1] = std::get<K-1>(t);
  }
};

template<class First, class Tuple, std::size_t N>
struct ArrayFiller<First, Tuple, N, 1> {
  static void fill_array_from_tuple( const Tuple& t, std::array<First, N> & arr) {
    arr[0] = std::get<0>(t);
  }
};

template<typename First, typename... Rem>
void fill_array_from_tuple(const std::tuple<First, Rem...>& t, std::array<First, 1+sizeof...(Rem)> & arr) {
  ArrayFiller<First, decltype(t), 1+sizeof...(Rem)>::fill_array_from_tuple(t, arr);
}

int main() {
  std::tuple<double, double, double> tup(0.1, 0.2, 0.3);
  std::array<double, 3> arr;
  fill_array_from_tuple(tup, arr);

  for (double x : arr)
    std::cout << x << " ";
  return 0;
}

#5

Even if title says C++11 I think C++14 solution is worth sharing (since everyone searching for problem will come up here anyway). This one can be used in compile time (constexpr proper) and much shorter than other solutions.

即使题目说c++ 11，我认为c++ 14解决方案还是值得分享的(因为每个搜索问题的人都会出现在这里)。这个可以在编译时(constexpr)中使用，并且比其他解决方案要短得多。

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

// Convert tuple into a array implementation
template<typename T, std::size_t N, typename Tuple,  std::size_t... I>
constexpr decltype(auto) t2a_impl(const Tuple& a, std::index_sequence<I...>)
{
        return std::array<T,N>{std::get<I>(a)...};
}

// Convert tuple into a array
template<typename Head, typename... T>
constexpr decltype(auto) t2a(const std::tuple<Head, T...>& a)
{
        using Tuple = std::tuple<Head, T...>;
        constexpr auto N = sizeof...(T) + 1;
        return t2a_impl<Head, N, Tuple>(a, std::make_index_sequence<N>());
}

int main()
{
    constexpr auto tuple = std::make_tuple(-1.3,2.1,3.5);
    constexpr auto array = t2a(tuple);

    static_assert(array.size() == 3, "err");

    for(auto k : array)
        std::cout << k << ' ';
    return 0;
}

Example

例子