std::indirect_result_t - cppreference.com
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template< class F, class... Is > requires (std::indirectly_readable<Is> && ...) && std::invocable<F, std::iter_reference_t<Is>...> using indirect_result_t = std::invoke_result_t<F, std::iter_reference_t<Is>...>; |
(C++20 起) | |
别名模板 indirect_result_t 获得在解引用 indirectly_readable 类型 Is... 的结果上调用 invocable 类型 F 的结果类型。
模板形参
| F | - | 可调用类型 |
| Is | - | 解引用到实参的间接可读类型 |
示例
#include <iterator> #include <type_traits> struct Fn { long operator()(const int&); int operator()(int&&); short operator()(int, int) const; auto operator()(const float) -> int&; void operator()(int[8]); }; static_assert( std::is_same_v<std::indirect_result_t<Fn, const int*>, long> and std::is_same_v<std::indirect_result_t<Fn, std::move_iterator<int*>>, int> and std::is_same_v<std::indirect_result_t<const Fn, int*, int*>, short> and std::is_same_v<std::indirect_result_t<Fn, float*>, int&> and std::is_same_v<std::indirect_result_t<Fn, int**>, void> ); int main() {}