std::experimental::make_array
来自cppreference.com
< cpp | experimental
定义于头文件 <experimental/array>
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template <class D = void, class... Types> constexpr std::array<VT /* see below */, sizeof...(Types)> make_array(Types&&... t); |
(库基础 TS v2) | |
创建一个 std::array ,其大小等于参数数量,且其元素从对应参数初始化。返回 std::array<VT, sizeof...(Types)>{std::forward<Types>(t)...}
若 D
为 void ,则推出的类型 VT
为 std::common_type_t<Types...> 。否则,类型为 D
。
若 D
为 void ,且 std::decay_t<Types>... 中任意一者是 std::reference_wrapper 的特化,则程序为病态。
可能的实现
namespace details { template<class> struct is_ref_wrapper : std::false_type {}; template<class T> struct is_ref_wrapper<std::reference_wrapper<T>> : std::true_type {}; template<class T> using not_ref_wrapper = std::experimental::negation<is_ref_wrapper<std::decay_t<T>>>; template <class D, class...> struct return_type_helper { using type = D; }; template <class... Types> struct return_type_helper<void, Types...> : std::common_type<Types...> { static_assert(std::experimental::conjunction_v<not_ref_wrapper<Types>...>, "Types cannot contain reference_wrappers when D is void"); }; template <class D, class... Types> using return_type = std::array<typename return_type_helper<D, Types...>::type, sizeof...(Types)>; } template < class D = void, class... Types> constexpr details::return_type<D, Types...> make_array(Types&&... t) { return {std::forward<Types>(t)... }; } |
示例
运行此代码
#include <experimental/array> #include <iostream> #include <type_traits> int main() { decltype(auto) arr = std::experimental::make_array(1, 2, 3, 4, 5); bool is_array_of_5_ints = std::is_same<decltype(arr), std::array<int, 5>>::value; std::cout << "Returns an array of five ints? "; std::cout << std::boolalpha << is_array_of_5_ints << '\n'; }
输出:
Returns an array of five ints? true
参阅
从内建数组创建 std::array 对象 (函数模板) |