NAME¶

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)
constexpr explicit 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 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());.

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);

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:¶

2020-11-15 was 1014 day(s) ago
2120-11-15 is 35510 day(s) from now
2023-08-26 is today!