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std::chrono::year_month_day::ok

來自 cppreference.com
 
C++
 
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chrono I/O
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std::chrono::year_month_day
 
constexpr bool ok() const noexcept;
 (C++20 起)

檢查此 year_month_day 物件是否表示有效的日曆日期。

[編輯] 返回值

如果此 year_month_day 物件表示有效的日曆日期，即儲存的年、月和日值均有效，且儲存的日值在給定年和月的日期範圍內，則返回 true。否則返回 false。

[編輯] 可能的實現

constexpr bool std::chrono::year_month_day::ok() const noexcept
{
    return year().ok() && month().ok() && day().ok() &&
        day() <= (year()/month()/std::chrono::last).day();
}

[編輯] 示例

執行此程式碼
#include <chrono>
 
int main()
{
    constexpr auto ymd1 {std::chrono::day(1)/std::chrono::July/2020};
    static_assert(ymd1.ok());
 
    constexpr auto ymd2 {std::chrono::year(2020)/7/42};
    static_assert(not ymd2.ok());
 
    constexpr auto ymd3 {std::chrono::February/29/2020}; // ok, leap year
    static_assert(ymd3.ok());
 
    constexpr auto ymd4 = ymd3 + std::chrono::years{1}; // bad, not a leap year
    static_assert(ymd4 == std::chrono::February/29/2021 and not ymd4.ok());
 
    // to fix the bad date we may want to snap to the last day of the month:
    if constexpr (!ymd4.ok())
    {
        constexpr auto ymd = ymd4.year()/ymd4.month()/std::chrono::last;
        static_assert(ymd == std::chrono::February/28/2021 and ymd.ok());
    }
 
    // or we may want to overflow to the next month:
    if constexpr (!ymd4.ok())
    {
        constexpr auto st = std::chrono::sys_time<std::chrono::days>{ymd4};
        constexpr auto ymd = std::chrono::year_month_day{st};
        static_assert(ymd == std::chrono::March/1/2021 and ymd.ok());
    }
}
取自“https://cppreference.tw/mwiki/index.php?title=cpp/chrono/year_month_day/ok&oldid=161691”