std::chrono::year_month_day::operatorsys_days,std::chrono::year_month_day::operator(3)

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!