std::chrono::year::is_leap

 

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std::chrono::year

成员函数

year::year

year::operator++year::operator--

year::operator+=year::operator-=

year::operator+year::operator-

year::is_leap

year::operator int

year::ok

year::min

year::max

非成员函数

operator==operator<=>

operator+operator-

operator<<

from_stream

operator""y

辅助类

formatter<std::chrono::year>

hash<std::chrono::year> (C++26)

 

constexpr bool is_leap() const noexcept;		(自 C++20 起)

确定 *this 在 推算的格里高利历 中是否代表闰年。

*this 代表闰年，如果存储的年份值

• 可以被 4 整除但不能被 100 整除；或者
• 可以被 400 整除。

返回值

如果 *this 代表闰年，则为 true，否则为 false。

示例

运行此代码

#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";
    }
}

输出

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