std::chrono::year::is_leap
來自 cppreference.com
constexpr bool is_leap() const noexcept; |
(C++20 起) | |
確定 *this 是否在 純粹格里高利曆 中表示閏年。
*this 表示閏年,如果儲存的年份值
- 能被 4 整除但不能被 100 整除;或
- 能被 400 整除。
[編輯] 返回值
true 如果 *this 表示閏年,否則 false。
[編輯] 示例
執行此程式碼
#include <chrono> #include <iostream> int main() { using namespace std::chrono_literals; for (const std::chrono::year y : {2020y, 2021y, 2000y, 3000y}) { if (const int iy{static_cast<int>(y)}; y.is_leap()) std::cout << iy << " is a leap year because it is divisible by " << (iy % 400 == 0 ? "400\n" : "4 and not divisible by 100\n"); else std::cout << iy << " is not a leap year\n"; } }
輸出
2020 is a leap year because it is divisible by 4 and not divisible by 100 2021 is not a leap year 2000 is a leap year because it is divisible by 400 3000 is not a leap year