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std::chrono::year_month_day::year_month_day
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(1) | (since C++20) |
|
(2) | (since C++20) |
|
(3) | (since C++20) |
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(4) | (since C++20) |
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(5) | (since C++20) |
Constructs a year_month_day object.
year_month_day object that stores the year y, month m and day d.
year_month_day object that stores the year ymdl.year(), month ymdl.month() and day ymdl.day().
year_month_day object that represent the same date as the one represented by dp.
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
#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
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(C++20)
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conventional syntax for Gregorian calendar date creation (function) |
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