constexpr operator std::chrono::sys_days() const noexcept; | (1) | (since C++20) |
explicit constexpr 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
.
ok()
is true, the return value holds a count of days from the std::chrono::system_clock
epoch (1970-01-01) to *this
. The result is negative if *this
represent a date prior to it.year().ok() && month().ok()
is true
), then the returned value is sys_days(year()/month()/1d) + (day() - 1d)
. year().ok() && month().ok()
is false
), the return value is unspecified.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.local_days
instead. Equivalent to return local_days(sys_days(*this).time_since_epoch());
.Converting to 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);
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