我有一个未知大小的元组（它是方法的模板参数）
是否可以获得它的一部分（我需要扔掉它的第一个元素）
例如，我有tuple<int,int,int>(7,12,42)。我想tuple<int,int>(12,42)在这里
答案 0 :(得分：15)
借助编译时整数列表：
#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;
};
我们可以通过参数包扩展来构造尾部：
#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 %g\n", s, d);
// prints: hello 7.9
return 0;
}
（关于ideone：http://ideone.com/Tzv7v;代码适用于g ++ 4.5到4.7和clang ++ 3.0）
答案 1 :(得分：9)
可能有一种更简单的方法，但这是一个开始。 “tail”函数模板返回一个复制的元组，其中包含除第一个之外的所有原始值。这在C ++ 0x模式下用GCC 4.6.2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);
}
答案 2 :(得分：8)
使用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 apply([](auto head, auto... tail) {
return std::make_tuple(tail...)};
}, t);
}
答案 3 :(得分：6)
我对Adam's code进行了一些修改，它将剥离元组的前N个参数，并创建一个只有最后N个类型的新元组......这是完整的代码（请注意：如果有人决定给我的答案+1，那么请点击Adam的答案，因为这是此代码的基础，我不希望从他的贡献中拿走任何信用）：
//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;
}
答案 4 :(得分：5)
元组切片操作（也适用于std::array和std::pair）可以像这样定义（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>{});
}
可以像这样获得元组t的任意子集：
tuple_slice<I1, I2>(t);
其中[I1, I2)是子集的独占范围，返回值是引用或值的元组，取决于输入元组是分别是左值还是右值（a我的blog可以找到详尽的阐述。
答案 5 :(得分：2)
请不要使用！
int，但可能会因您的类型而失败！）。请参阅评论以供讨论。我留下这个答案仅供参考。
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;
我得到了：
tail1: {12, 42}
tail2: {42}
我不确定这不是一个未指明的行为。适合我：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.
答案 6 :(得分：0)
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;
}