std::chrono::year_month_day:: year_month_day

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std::chrono::year_month_day
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.

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 / .

Example

Run this code 
#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

See also

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