std::chrono::year_month_day::operator sys_days, std::chrono::year_month_day::operator local_days
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std::chrono::year_month_day
Member functions
year_month_day::operator sys_daysyear_month_day::operator local_days
Nonmember functions
Helper classes
constexpr operator std::chrono::sys_days () const noexcept;
(1)
(since C++20)
constexpr explicit operator std::chrono::local_days () const noexcept;
(2)
(since C++20)
Converts *this to a std::chrono::time_point representing the same date as this year_month_day.
1) If ok() is true, the return value holds a count of days from the std::chrono::system_clock epoch (1970年01月01日) to *this. The result is negative if *this represent a date prior to it.
Otherwise, if the stored year and month are valid (year().ok() && month().ok() is true), then the returned value is sys_days(year()/month()/1d) + (day() - 1d).
Otherwise (if year().ok() && month().ok() is false), the return value is unspecified.
A std::chrono::sys_days in the range [std::chrono::days {-12687428}, std::chrono::days {11248737}], when converted to
year_month_day and back, yields the same value.2) Same as (1) but returns local_days instead. Equivalent to return local_days(sys_days(*this).time_since_epoch());.
[edit] Notes
Converting to std::chrono::sys_days and back can be used to normalize a year_month_day that contains an invalid day but a valid year and month:
using namespace std::chrono; auto ymd = 2017y/January/0; ymd = sys_days{ymd}; // ymd is now 2016y/December/31
Normalizing the year and month can be done by adding (or subtracting) zero std::chrono::months:
using namespace std::chrono; constexpr year_month_day normalize(year_month_day ymd) { ymd += months{0}; // normalizes year and month return sys_days{ymd}; // normalizes day } static_assert(normalize(2017y/33/59) == 2019y/10/29);
[edit] Example
Run this code
#include <chrono> #include <iostream> int main() { using namespace std::chrono; const auto today = sys_days{std::chrono::floor <days>(system_clock::now())}; for (const year_month_day ymd : {{November/15/2020}, {November/15/2120}, today}) { std::cout << ymd; const auto delta = (sys_days{ymd} - today).count(); (delta < 0) ? std::cout << " was " << -delta << " day(s) ago\n" : (delta > 0) ? std::cout << " is " << delta << " day(s) from now\n" : std::cout << " is today!\n"; } }
Possible output:
2020年11月15日 was 1014 day(s) ago 2120年11月15日 is 35510 day(s) from now 2023年08月26日 is today!