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std::chrono::year_month_day::year_month_day

From cppreference.com
 
 
Utilities library
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year_month_day() = default;
(1) (since C++20)
constexpr year_month_day( const std::chrono::year& y,

                          const std::chrono::month& m,

                          const std::chrono::day& d ) noexcept;
(2) (since C++20)
constexpr year_month_day( const std::chrono::year_month_day_last& ymdl ) noexcept;
(3) (since C++20)
constexpr year_month_day( const std::chrono::sys_days& dp ) noexcept;
(4) (since C++20)
constexpr explicit year_month_day( const std::chrono::local_days& dp ) noexcept;
(5) (since C++20)

Constructs a year_month_day object.

1) Default constructor leaves the date uninitialized.
2) Constructs a year_month_day object that stores the year y, month m and day d.
3) Constructs a year_month_day object that stores the year ymdl.year(), month ymdl.month() and day ymdl.day().
4) Constructs a year_month_day object that represent the same date as the one represented by dp.
5) Constructs a year_month_day object that represent the same date as the one represented by dp, as if by year_month_day(sys_days(dp.time_since_epoch())).

Constructors (3,4) define implicit conversions from std::chrono::year_month_day_last and std::chrono::sys_days, respectively.

For any year_month_day object ymd representing a valid date (ymd.ok() == true), converting ymd to sys_days and back yields the same value.

[edit] Notes

A year_month_day can also be created by combining one of the partial-date types std::chrono::year_month and std::chrono::month_day with the missing component (day and year, respectively) using operator/.

[edit] Example

#include <chrono>
#include <iostream>
 
int main()
{
    using namespace std::chrono;
 
    constexpr auto ymd2 = std::chrono::year_month_day(
        2020y, std::chrono::January, 31d // overload (2)
    );
    std::cout << "ymd2: " << ymd2 << '\n';
 
    constexpr auto ymd3 = std::chrono::year_month_day(
        std::chrono::April / std::chrono::Monday[last] / 2023y // overload (3)
    );
    std::cout << "ymd3: " << ymd3 << '\n';
 
    const auto now = std::chrono::system_clock::now();
    const auto ymd4 = std::chrono::year_month_day(
        std::chrono::floor<std::chrono::days>(now) // overload (4)
    );
    std::cout << "ymd4: " << ymd4 << '\n';
}

Possible output:

ymd2: 2020-01-31
ymd3: 2023-04-24
ymd4: 2023-08-30

[edit] See also

(C++20)
conventional syntax for Gregorian calendar date creation
(function) [edit]