std::chrono::year_month_day::operator sys_days, std::chrono::year_month_day::operator local_days

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C-style date and time
 
std::chrono::year_month_day
 
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.

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

Run this code
#include <iostream>
#include <chrono>
using namespace std::chrono;
 
int main()
{
    constexpr auto ymd {November/15/2021};
    constexpr auto sd = sys_days{ymd};
    auto today = sys_days{floor<days>(system_clock::now())};
    auto delta = (sd - today).count();
    std::cout << "November 15, 2021 ";
    if (delta < 0) 
        std::cout << "was " << -delta << " day(s) ago.";
    else if (delta == 0)
        std::cout << "is today!";
    else 
        std::cout << "is " << delta << " day(s) from now.";
    std::cout << '\n';
}

Possible output: 

November 15, 2021 is 437 day(s) from now.
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