std::chrono::year_month_day_last:: operator sys_days, std::chrono::year_month_day_last:: operator local_days

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C-style date and time
std::chrono::year_month_day_last
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_last . This is equivalent to composing a year_month_day from year() , month() and day() and converting that year_month_day to the destination type.

1) Equivalent to std:: chrono :: sys_days ( year ( ) / month ( ) / day ( ) ) .
2) Equivalent to std:: chrono :: local_days ( year ( ) / month ( ) / day ( ) ) .

Example

Run this code 
#include <array>
#include <chrono>
#include <iostream>
#include <string_view>
using namespace std::chrono;
using namespace std::literals;
 
int main()
{
    constexpr std::chrono::year y{2023y};
    constexpr std::array quarters{"1st"sv, "2nd"sv, "3rd"sv, "4th"sv};
    constexpr auto mq{12 / 4}; // months per quarter        
 
    std::cout << "In year " << static_cast<int>(y) << '\n';
    for (auto q = 1; q < 5; ++q)
    {
        const auto ls = y / std::chrono::month(q * mq) / Sunday[last];
        const auto ld = y / std::chrono::month(q * mq) / last;
        // subtract last Sunday from last day for day of week
        const auto index = (sys_days(ld) - sys_days(ls)).count();
        std::cout << "The " << quarters[q - 1] << " quarter ends on a "  
                  << std::chrono::weekday(index) << '\n';
    }
}

Output: 

In year 2023
The 1st quarter ends on a Fri
The 2nd quarter ends on a Fri
The 3rd quarter ends on a Sat
The 4th quarter ends on a Sun

See also

converts to a std::chrono::time_point
(public member function of std::chrono::year_month_day )
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