std::chrono::year_month_day::operatorsys_days,std::chrono::year_month_day::operator (3) Linux Manual Page

std::chrono::year_month_day::operatorsys_days,std::chrono::year_month_day::operator – std::chrono::year_month_day::operatorsys_days,std::chrono::year_month_day::operator

Synopsis

constexpr operator std::chrono::sys_days() const noexcept;
(1)(since C++ 20)

    explicit constexpr
    operator std::chrono::local_days() const(2)(since C++ 20)

        noexcept;

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 (1970-01-01) 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 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());
.

Notes

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);

Example

 This section is incomplete
 Reason: no example