获取std::元组的一部分

在编译时整数列表的帮助下:

#include <cstdlib>
template <size_t... n>
struct ct_integers_list {
    template <size_t m>
    struct push_back
    {
        typedef ct_integers_list<n..., m> type;
    };
};
template <size_t max>
struct ct_iota_1
{
    typedef typename ct_iota_1<max-1>::type::template push_back<max>::type type;
};
template <>
struct ct_iota_1<0>
{
    typedef ct_integers_list<> type;
};

我们可以简单地通过参数包展开来构造尾部:

#include <tuple>
template <size_t... indices, typename Tuple>
auto tuple_subset(const Tuple& tpl, ct_integers_list<indices...>)
    -> decltype(std::make_tuple(std::get<indices>(tpl)...))
{
    return std::make_tuple(std::get<indices>(tpl)...);
    // this means:
    //   make_tuple(get<indices[0]>(tpl), get<indices[1]>(tpl), ...)
}
template <typename Head, typename... Tail>
std::tuple<Tail...> tuple_tail(const std::tuple<Head, Tail...>& tpl)
{
    return tuple_subset(tpl, typename ct_iota_1<sizeof...(Tail)>::type());
    // this means:
    //   tuple_subset<1, 2, 3, ..., sizeof...(Tail)-1>(tpl, ..)
}

用法:

#include <cstdio>
int main()
{
    auto a = std::make_tuple(1, "hello", 7.9);
    auto b = tuple_tail(a);
    const char* s = nullptr;
    double d = 0.0;
    std::tie(s, d) = b;
    printf("%s %gn", s, d);
    // prints:   hello 7.9
    return 0;
}

(On ideone: http://ideone.com/Tzv7v;该代码适用于g++ 4.5到4.7和clang++ 3.0)

对于c++ 17，您可以使用std::apply:

template <typename Head, typename... Tail>
std::tuple<Tail...> tuple_tail(const std::tuple<Head, Tail...>& t)
{
    return std::apply([](auto head, auto... tail) {
        return std::make_tuple(tail...);
    }, t);
}

可能有更简单的方法，但这是一个开始。"tail"函数模板返回一个复制的元组，其中包含原始元组的所有值，除了第一个。在c++ 0模式下使用GCC 4.6.2进行编译。

template<size_t I>
struct assign {
  template<class ResultTuple, class SrcTuple>
  static void x(ResultTuple& t, const SrcTuple& tup) {
    std::get<I - 1>(t) = std::get<I>(tup);
    assign<I - 1>::x(t, tup);
  }
};
template<>
struct assign<1> {
  template<class ResultTuple, class SrcTuple>
  static void x(ResultTuple& t, const SrcTuple& tup) {
    std::get<0>(t) = std::get<1>(tup);
  }
};

template<class Tup> struct tail_helper;
template<class Head, class... Tail>
struct tail_helper<std::tuple<Head, Tail...>> {
  typedef typename std::tuple<Tail...> type;
  static type tail(const std::tuple<Head, Tail...>& tup) {
    type t;
    assign<std::tuple_size<type>::value>::x(t, tup);
    return t;
  }
};
template<class Tup>
typename tail_helper<Tup>::type tail(const Tup& tup) {
  return tail_helper<Tup>::tail(tup);
}

元组切片操作(也适用于std::array和std::pair)可以这样定义(需要c++ 14):

namespace detail
{
    template <std::size_t Ofst, class Tuple, std::size_t... I>
    constexpr auto slice_impl(Tuple&& t, std::index_sequence<I...>)
    {
        return std::forward_as_tuple(
            std::get<I + Ofst>(std::forward<Tuple>(t))...);
    }
}
template <std::size_t I1, std::size_t I2, class Cont>
constexpr auto tuple_slice(Cont&& t)
{
    static_assert(I2 >= I1, "invalid slice");
    static_assert(std::tuple_size<std::decay_t<Cont>>::value >= I2, 
        "slice index out of bounds");
    return detail::slice_impl<I1>(std::forward<Cont>(t),
        std::make_index_sequence<I2 - I1>{});
}

可以得到元组t的任意子集，如下所示:

tuple_slice<I1, I2>(t); 

其中[I1, I2)是子集的独占范围，返回值是对原始元组的引用的元组。如果需要复制切片操作，将slice_impl中的make_tuple替换为forward_as_tuple。

我对Adam的代码做了一些修改，去掉元组的前N个参数，并创建一个只包含最后N个类型的新元组…这里是完整的代码(注意:如果有人决定+1我的答案，也请+1亚当的答案，因为这是这个代码的基础，我不希望从他的贡献中拿走任何荣誉):

//create a struct that allows us to create a new tupe-type with the first
//N types truncated from the front
template<size_t N, typename Tuple_Type>
struct tuple_trunc {};
template<size_t N, typename Head, typename... Tail>
struct tuple_trunc<N, std::tuple<Head, Tail...>>
{
    typedef typename tuple_trunc<N-1, std::tuple<Tail...>>::type type;
};
template<typename Head, typename... Tail>
struct tuple_trunc<0, std::tuple<Head, Tail...>>
{
    typedef std::tuple<Head, Tail...> type;
};
/*-------Begin Adam's Code-----------
Note the code has been slightly modified ... I didn't see the need for the extra
variadic templates in the "assign" structure.  Hopefully this doesn't break something
I didn't forsee
*/
template<size_t N, size_t I>
struct assign 
{
    template<class ResultTuple, class SrcTuple>
    static void x(ResultTuple& t, const SrcTuple& tup) 
    {
        std::get<I - N>(t) = std::get<I>(tup);  
        assign<N, I - 1>::x(t, tup);  //this offsets the assignment index by N
    }
};
template<size_t N>
struct assign<N, 1> 
{
    template<class ResultTuple, class SrcTuple>
    static void x(ResultTuple& t, const SrcTuple& tup) 
    {
        std::get<0>(t) = std::get<1>(tup);
    }
};

template<size_t TruncSize, class Tup> struct th2;
//modifications to this class change "type" to the new truncated tuple type
//as well as modifying the template arguments to assign
template<size_t TruncSize, class Head, class... Tail>
struct th2<TruncSize, std::tuple<Head, Tail...>> 
{
    typedef typename tuple_trunc<TruncSize, std::tuple<Head, Tail...>>::type type;
    static type tail(const std::tuple<Head, Tail...>& tup) 
    {
        type t;
        assign<TruncSize, std::tuple_size<type>::value>::x(t, tup);
        return t;
    }
};
template<size_t TruncSize, class Tup>
typename th2<TruncSize, Tup>::type tail(const Tup& tup) 
{
    return th2<TruncSize, Tup>::tail(tup);
}
//a small example
int main()
{
    std::tuple<double, double, int, double> test(1, 2, 3, 4);
    tuple_trunc<2, std::tuple<double, double, int, double>>::type c = tail<2>(test);
    return 0;
}

请勿使用!

• 这是[很可能]未指定的行为。可以随时停止工作。
• 此外，还有填充问题的可能性(即可能适用于int，但可能不适用于您的类型!)。

见注释进行讨论。我把这个答案留作参考。

更简单:

tuple<int,int,int> origin{7,12,42};
tuple<int, int> &tail1 = (tuple<int, int>&)origin;
tuple<int> &tail2 = (tuple<int>&)origin;
cout << "tail1: {" << get<0>(tail1) << ", " << get<1>(tail1) << "}" << endl;
cout << "tail2: {" << get<0>(tail2) << "}" << endl;

tail1: {12, 42}
tail2: {42}

我不确定这不是一个未指定的行为。适用于我:Fedora 20和

❯ clang --version
clang version 3.3 (tags/RELEASE_33/final)
Target: x86_64-redhat-linux-gnu
Thread model: posix
❯ gcc --version
gcc (GCC) 4.8.2 20131212 (Red Hat 4.8.2-7)
Copyright (C) 2013 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

参考资料:voidnish.wordpress.com/上的文章

c++17方式:

#include <tuple>
#include <string>
template <size_t __begin, size_t...__indices, typename Tuple>
auto tuple_slice_impl(Tuple &&tuple, std::index_sequence<__indices...>) {
  return std::make_tuple(std::get<__begin + __indices>(std::forward<Tuple>(tuple))...);
}
template <size_t __begin, size_t __count, typename Tuple>
auto tuple_slice(Tuple &&tuple) {
  static_assert(__count > 0, "splicing tuple to 0-length is weird...");
  return tuple_slice_impl<__begin>(std::forward<Tuple>(tuple), std::make_index_sequence<__count>());
}
template <size_t __begin, size_t __count, typename Tuple>
using tuple_slice_t = decltype(tuple_slice<__begin, __count>(Tuple{}));
using test_tuple = std::tuple<int, int, bool, nullptr_t, std::string>;
using sliced_test = tuple_slice_t<2, 2, test_tuple>;
static_assert(std::tuple_size_v<sliced_test> == 2);
static_assert(std::is_same_v<std::decay_t<decltype(std::get<0>(sliced_test{}))>, bool>);
static_assert(std::is_same_v<std::decay_t<decltype(std::get<1>(sliced_test{}))>, nullptr_t>);
#include <cassert>
int main() {
  test_tuple tuple {
    -1, 42, true, nullptr, "hello"
  };
  auto spliced = tuple_slice<3, 2>(tuple);
  assert(std::get<0>(spliced) == nullptr);
  assert(std::get<1>(spliced) == std::get<4>(tuple));
  return 0;
}