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

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