如何生成 N 类型 T 的元组?
How can I generate a tuple of N type T's?
我希望能够编写 generate_tuple_type<int, 3>
,它在内部具有类型别名 type
,在这种情况下为 std::tuple<int, int, int>
。
一些示例用法:
int main()
{
using gen_tuple_t = generate_tuple_type<int, 3>::type;
using hand_tuple_t = std::tuple<int, int, int>;
static_assert( std::is_same<gen_tuple_t, hand_tuple_t>::value, "different types" );
}
我怎样才能做到这一点?
检查此 link 的底部以获取示例:
http://en.cppreference.com/w/cpp/utility/integer_sequence.
您需要做更多的工作来将生成的元组封装为类型别名,但这里的关键构造是 std::integer_sequence
和朋友。
相当简单的递归公式:
template<typename T, unsigned N, typename... REST>
struct generate_tuple_type
{
typedef typename generate_tuple_type<T, N-1, T, REST...>::type type;
};
template<typename T, typename... REST>
struct generate_tuple_type<T, 0, REST...>
{
typedef std::tuple<REST...> type;
};
[更新]
好的,所以我只是在考虑 N
的适度值。以下公式更复杂,但也明显更快,并且对大参数的编译器破坏更少。
#include <tuple>
template<typename /*LEFT_TUPLE*/, typename /*RIGHT_TUPLE*/>
struct join_tuples
{
};
template<typename... LEFT, typename... RIGHT>
struct join_tuples<std::tuple<LEFT...>, std::tuple<RIGHT...>>
{
typedef std::tuple<LEFT..., RIGHT...> type;
};
template<typename T, unsigned N>
struct generate_tuple_type
{
typedef typename generate_tuple_type<T, N/2>::type left;
typedef typename generate_tuple_type<T, N/2 + N%2>::type right;
typedef typename join_tuples<left, right>::type type;
};
template<typename T>
struct generate_tuple_type<T, 1>
{
typedef std::tuple<T> type;
};
template<typename T>
struct generate_tuple_type<T, 0>
{
typedef std::tuple<> type;
};
int main()
{
using gen_tuple_t = generate_tuple_type<int, 30000>::type;
static_assert( std::tuple_size<gen_tuple_t>::value == 30000, "wrong size" );
}
这个版本最多执行 2*log(N)+1 个模板实例化,假设你的编译器记住了它们。证明留作 reader.
的练习
你可以用std::make_index_sequence
给你一个足够长的包,然后把它包成你需要的类型就行了。不需要递归:
template <typename T, size_t N>
class generate_tuple_type {
template <typename = std::make_index_sequence<N>>
struct impl;
template <size_t... Is>
struct impl<std::index_sequence<Is...>> {
template <size_t >
using wrap = T;
using type = std::tuple<wrap<Is>...>;
};
public:
using type = typename impl<>::type;
};
这是带有 Boost.Mp11 的 simple one liner。
#include <boost/mp11/algorithm.hpp>
int main()
{
using gen_tuple_t = boost::mp11::mp_repeat_c<std::tuple<int>, 3>;
using hand_tuple_t = std::tuple<int, int, int>;
static_assert( std::is_same_v<gen_tuple_t, hand_tuple_t>, "different types" );
}
我希望能够编写 generate_tuple_type<int, 3>
,它在内部具有类型别名 type
,在这种情况下为 std::tuple<int, int, int>
。
一些示例用法:
int main()
{
using gen_tuple_t = generate_tuple_type<int, 3>::type;
using hand_tuple_t = std::tuple<int, int, int>;
static_assert( std::is_same<gen_tuple_t, hand_tuple_t>::value, "different types" );
}
我怎样才能做到这一点?
检查此 link 的底部以获取示例:
http://en.cppreference.com/w/cpp/utility/integer_sequence.
您需要做更多的工作来将生成的元组封装为类型别名,但这里的关键构造是 std::integer_sequence
和朋友。
相当简单的递归公式:
template<typename T, unsigned N, typename... REST>
struct generate_tuple_type
{
typedef typename generate_tuple_type<T, N-1, T, REST...>::type type;
};
template<typename T, typename... REST>
struct generate_tuple_type<T, 0, REST...>
{
typedef std::tuple<REST...> type;
};
[更新]
好的,所以我只是在考虑 N
的适度值。以下公式更复杂,但也明显更快,并且对大参数的编译器破坏更少。
#include <tuple>
template<typename /*LEFT_TUPLE*/, typename /*RIGHT_TUPLE*/>
struct join_tuples
{
};
template<typename... LEFT, typename... RIGHT>
struct join_tuples<std::tuple<LEFT...>, std::tuple<RIGHT...>>
{
typedef std::tuple<LEFT..., RIGHT...> type;
};
template<typename T, unsigned N>
struct generate_tuple_type
{
typedef typename generate_tuple_type<T, N/2>::type left;
typedef typename generate_tuple_type<T, N/2 + N%2>::type right;
typedef typename join_tuples<left, right>::type type;
};
template<typename T>
struct generate_tuple_type<T, 1>
{
typedef std::tuple<T> type;
};
template<typename T>
struct generate_tuple_type<T, 0>
{
typedef std::tuple<> type;
};
int main()
{
using gen_tuple_t = generate_tuple_type<int, 30000>::type;
static_assert( std::tuple_size<gen_tuple_t>::value == 30000, "wrong size" );
}
这个版本最多执行 2*log(N)+1 个模板实例化,假设你的编译器记住了它们。证明留作 reader.
的练习你可以用std::make_index_sequence
给你一个足够长的包,然后把它包成你需要的类型就行了。不需要递归:
template <typename T, size_t N>
class generate_tuple_type {
template <typename = std::make_index_sequence<N>>
struct impl;
template <size_t... Is>
struct impl<std::index_sequence<Is...>> {
template <size_t >
using wrap = T;
using type = std::tuple<wrap<Is>...>;
};
public:
using type = typename impl<>::type;
};
这是带有 Boost.Mp11 的 simple one liner。
#include <boost/mp11/algorithm.hpp>
int main()
{
using gen_tuple_t = boost::mp11::mp_repeat_c<std::tuple<int>, 3>;
using hand_tuple_t = std::tuple<int, int, int>;
static_assert( std::is_same_v<gen_tuple_t, hand_tuple_t>, "different types" );
}