C++17将会有make_from_tuple
这样做,但你可以用 C++11 编写它。这是盗自的 C++14 版本参考参数 http://en.cppreference.com/w/cpp/utility/make_from_tuple(对于 C++11,您可以使用以下实现std::index_sequence
from 实现 C++14 make_integer_sequence https://stackoverflow.com/questions/17424477/implementation-c14-make-integer-sequence).
namespace detail {
template <class T, class Tuple, std::size_t... I>
constexpr T make_from_tuple_impl( Tuple&& t, std::index_sequence<I...> )
{
return T(std::get<I>(std::forward<Tuple>(t))...);
}
} // namespace detail
template <class T, class Tuple>
constexpr T make_from_tuple( Tuple&& t )
{
return detail::make_from_tuple_impl<T>(std::forward<Tuple>(t),
std::make_index_sequence<std::tuple_size<std::decay_t<Tuple>>::value>{});
}
有了这个实用程序,实施f
轻而易举:
template <typename... Args0, typename... Args1>
void f(std::tuple<Args0...> args0, std::tuple<Args1...> args1) {
auto alpha = make_from_tuple<Object0>(args0);
auto beta = make_from_tuple<Object1>(args1);
}
为了使其更加通用,我建议仅推断这些元组的类型并完美转发它们:
template <typename T0, typename T1>
void f(T0&& args0, T1&& args1) {
auto alpha = make_from_tuple<Object0>(std::forward<T0>(args0));
auto beta = make_from_tuple<Object1>(std::forward<T1>(args1));
}
现场 C++11 演示 http://coliru.stacked-crooked.com/a/859b956248e783a9