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std::chrono...ar_month_day(3) C++ Standard Libarystd::chrono...ar_month_day(3)

NAME
       std::chrono::year_month_day::year_month_day			     -
       std::chrono::year_month_day::year_month_day

Synopsis
	  year_month_day()		       =		      default;
       (1) (since C++20)
	  constexpr year_month_day( const std::chrono::year& y,

				    const	 std::chrono::month&	    m,
       (2) (since C++20)

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

	  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  par-
       tial-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
	  operator/ conventional syntax	for Gregorian calendar date creation
	  (C++20)   (function)

http://cppreference.com		  2024.06.10	 std::chrono...ar_month_day(3)

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