C++ 从std:：tuple开始创建子元组

让我们假设给定了一个std:：tuple。我想创建一个新的std:：tuple，其类型是[0，sizeof…some_types-2]中索引的类型。例如，假设起始元组是std:：tuple。我想获得定义为std:：tuple的子元组

我对可变模板很陌生。作为第一步，我尝试编写一个负责存储原始std:：tuple的不同类型的结构，目的是创建一个与std:：tuple new_tuple中的类型相同的新元组

我想做的是：

std::tuple<type_list<bool, double, int>::type...> new_tuple // this won't work

下一步是丢弃参数包中的最后一个元素。如何访问类型列表中存储的多个类型？如何丢弃其中的一些

---

谢谢。

一种方法是递归地将两个元组传递给helper结构，该结构接受源元组的第一个元素，并将其添加到另一个元组的末尾：

#include <iostream>
#include <tuple>
#include <type_traits>

namespace detail {

    template<typename...>
    struct truncate;

    // this specialization does the majority of the work

    template<typename... Head, typename T, typename... Tail>
    struct truncate< std::tuple<Head...>, std::tuple<T, Tail...> > {
        typedef typename
        truncate< std::tuple<Head..., T>, std::tuple<Tail...> >::type type;
    };

    // this one stops the recursion when there's only
    // one element left in the source tuple

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

template<typename...>
struct tuple_truncate;

template<typename... Args>
struct tuple_truncate<std::tuple<Args...>> {

    // initiate the recursion - we start with an empty tuple,
    // with the source tuple on the right

    typedef typename detail::truncate< std::tuple<>, std::tuple<Args...> >::type type;
};

int main()
{
    typedef typename tuple_truncate< std::tuple<bool, double, int> >::type X;

    // test
    std::cout << std::is_same<X, std::tuple<bool, double>>::value; // 1, yay
}

---

.

使用索引序列技术，这种操作相当容易：生成一个索引序列，比元组少两个索引，并使用该序列从原始字段中选择字段。从C++14使用和返回类型推断：

template <typename... T, std::size_t... I>
auto subtuple_(const std::tuple<T...>& t, std::index_sequence<I...>) {
  return std::make_tuple(std::get<I>(t)...);
}

template <int Trim, typename... T>
auto subtuple(const std::tuple<T...>& t) {
  return subtuple_(t, std::make_index_sequence<sizeof...(T) - Trim>());
}

在C++11中：

#include <cstddef>     // for std::size_t

template<typename T, T... I>
struct integer_sequence {
  using value_type = T;

  static constexpr std::size_t size() noexcept {
    return sizeof...(I);
  }
};

namespace integer_sequence_detail {
template <typename, typename> struct concat;

template <typename T, T... A, T... B>
struct concat<integer_sequence<T, A...>, integer_sequence<T, B...>> {
  typedef integer_sequence<T, A..., B...> type;
};

template <typename T, int First, int Count>
struct build_helper {
  using type = typename concat<
    typename build_helper<T, First,           Count/2>::type,
    typename build_helper<T, First + Count/2, Count - Count/2>::type
  >::type;
};

template <typename T, int First>
struct build_helper<T, First, 1> {
  using type = integer_sequence<T, T(First)>;
};

template <typename T, int First>
struct build_helper<T, First, 0> {
  using type = integer_sequence<T>;
};

template <typename T, T N>
using builder = typename build_helper<T, 0, N>::type;
} // namespace integer_sequence_detail

template <typename T, T N>
using make_integer_sequence = integer_sequence_detail::builder<T, N>;

template <std::size_t... I>
using index_sequence = integer_sequence<std::size_t, I...>;

template<size_t N>
using make_index_sequence = make_integer_sequence<size_t, N>;

#include <tuple>

template <typename... T, std::size_t... I>
auto subtuple_(const std::tuple<T...>& t, index_sequence<I...>) 
  -> decltype(std::make_tuple(std::get<I>(t)...))
{
  return std::make_tuple(std::get<I>(t)...);
}

template <int Trim, typename... T>
auto subtuple(const std::tuple<T...>& t)
  -> decltype(subtuple_(t, make_index_sequence<sizeof...(T) - Trim>()))
{
  return subtuple_(t, make_index_sequence<sizeof...(T) - Trim>());
}

---

.

这里有一种直接解决问题的方法

template<unsigned...s> struct seq { typedef seq<s...> type; };
template<unsigned max, unsigned... s> struct make_seq:make_seq<max-1, max-1, s...> {};
template<unsigned...s> struct make_seq<0, s...>:seq<s...> {};

template<unsigned... s, typename Tuple>
auto extract_tuple( seq<s...>, Tuple& tup ) {
  return std::make_tuple( std::get<s>(tup)... );
}

---

这为您提供了一种将type_列表的类型累积到任意参数包保存模板中的方法。但您的问题不需要这样做。

具有边界检查的元组的子范围，无需声明帮助器类：

template <size_t starting, size_t elems, class tuple, class seq = decltype(std::make_index_sequence<elems>())>
struct sub_range;

template <size_t starting, size_t elems, class ... args, size_t ... indx>
struct sub_range<starting, elems, std::tuple<args...>, std::index_sequence<indx...>>
{
    static_assert(elems <= sizeof...(args) - starting, "sub range is out of bounds!");
    using tuple = std::tuple<std::tuple_element_t<indx + starting, std::tuple<args...>> ...>;
};

用法：

struct a0;
...
struct a8;

using range_outer = std::tuple<a0, a1, a2, a3, a4, a5, a6, a7, a8>;
sub_range<2, 3, range_outer>::tuple; //std::tuple<a2, a3, a4>

---

您的意思是[0，sizeof…某些_类型-2]的可能重复，即您的示例中的[0，1]？是的，已更正。谢谢，比你早两分钟@凯西矿只有一半的长度。。。如果包含符合C++14的整数_序列的副本；OHH，我在MaxySeq中有指数递归，这很好，对吗？FWWW，我认为log n索引对于简单的Script片段来说太麻烦了，所以对于清晰简洁的版本，+1到YAKK。P@Yakk我很喜欢你的抽取元组的泛型。泛型是一个词吗？我该睡觉了。@Yakk现在我休息得更好了，请允许我注意，1我甚至没有注意到我的双entendre是Yakk之前长度的一半，不小于，2在执行extract\u tuple时忘记了扩展s。固定的

template<template<typename...>class target>
struct gather {
  typedef typename type_list<types...>::template gather<target>::type parent_result;
  typedef typename append< parent_result, T >::type type;
};

template <size_t starting, size_t elems, class tuple, class seq = decltype(std::make_index_sequence<elems>())>
struct sub_range;

template <size_t starting, size_t elems, class ... args, size_t ... indx>
struct sub_range<starting, elems, std::tuple<args...>, std::index_sequence<indx...>>
{
    static_assert(elems <= sizeof...(args) - starting, "sub range is out of bounds!");
    using tuple = std::tuple<std::tuple_element_t<indx + starting, std::tuple<args...>> ...>;
};

struct a0;
...
struct a8;

using range_outer = std::tuple<a0, a1, a2, a3, a4, a5, a6, a7, a8>;
sub_range<2, 3, range_outer>::tuple; //std::tuple<a2, a3, a4>