我有下一个代码:
std::cout << (-10 >> 1) << std::endl;
std::cout << (-9 >> 1) << std::endl;
std::cout << (-8 >> 1) << std::endl;
std::cout << (-7 >> 1) << std::endl;
std::cout << (-6 >> 1) << std::endl;
std::cout << (-5 >> 1) << std::endl;
std::cout << (-4 >> 1) << std::endl;
std::cout << (-3 >> 1) << std::endl;
std::cout << (-2 >> 1) << std::endl;
std::cout << (-1 >> 1) << std::endl;
结果是:
-5
-5
-4
-4
-3
-3
-2
-2 …如何将类型从可变参数模板参数转换为另一种类型？
例如:
template <typename... T>
struct single
{
std::tuple<T...> m_single;
};
template <typename... T>
struct sequences
{
single<T...> get(size_t pos)
{
// I don't know how to convert here
return std::make_tuple(std::get<0>(m_sequences)[pos]... std::get<N>(m_sequences)[pos]);
}
template <size_t Idx>
std::vector<
typename std::tuple_element<Idx, std::tuple<T...>>::type
>
get_sequence()
{
return std::get<Idx>(m_sequences);
}
std::tuple<T...> m_sequences; // std::tuple<std::vector<T...>> I don't know how to conver here
};
我想这样写:
sequences<int, double, double> seq;
single<int, double, double> sin = seq.get(10);
并且具有std::tuple<std::vector<int>, std::vector<double>, std::vector<double>>结构序列.并从中获得单身.
std::vector<single<T...>> 对我来说是个坏主意,因为我需要一个完整的序列,并且很容易从中复制它.
可能吗？
非常感谢你.对不起,我的英语不好.