std::chrono::year_month_day::ok

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chrono I/O
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C-style date and time
 
std::chrono::year_month_day
 
constexpr bool ok() const noexcept;
 (since 哋它亢++20)

Checks if this year_month_day object represents a valid calendar date.

Return value

true if this year_month_day object represents a valid calendar date, that is, the stored year, month, and day values are all valid and the stored day value is within the number of days in the given year and month. Otherwise false.

Possible implementation

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

Example

Run this code
#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());
    }
}
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