std::chrono::operator+, std::chrono::operator- (std::chrono::year_month_day_last)

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Defined in header <chrono>
constexpr std::chrono::year_month_day_last

operator+( const std::chrono::year_month_day_last& ymdl, const std::chrono::months& dm )

    noexcept;
(since C++20)
constexpr std::chrono::year_month_day_last

operator+( const std::chrono::months& dm, const std::chrono::year_month_day_last& ymdl )

    noexcept;
(since C++20)
constexpr std::chrono::year_month_day_last

operator+( const std::chrono::year_month_day_last& ymdl, const std::chrono::years& dy )

    noexcept;
(since C++20)
constexpr std::chrono::year_month_day_last

operator+( const std::chrono::years& dy, const std::chrono::year_month_day_last& ymdl )

    noexcept;
(since C++20)
constexpr std::chrono::year_month_day_last

operator-( const std::chrono::year_month_day_last& ymdl, const std::chrono::months& dm )

    noexcept;
(since C++20)
constexpr std::chrono::year_month_day_last

operator-( const std::chrono::year_month_day_last& ymdl, const std::chrono::years& dy )

    noexcept;
(since C++20)
1-2) Adds dm.count() months to the date represented by ymdl. The result has the same year() and month() as std::chrono::year_month(ymdl.year(), ymdl.month()) + dm.
3-4) Adds dy.count() years to the date represented by ymdl. The result is equivalent to std::chrono::year_month_day_last(ymdl.year() + dy, ymdl.month_day_last()).
5) Subtracts dm.count() months from the date represented by ymdl. Equivalent to ymdl + -dm.
6) Subtracts dy.count() years from the date represented by ymdl. Equivalent to ymdl + -dy.

For durations that are convertible to both std::chrono::years and std::chrono::months, the years overloads (3,4,6) are preferred if the call would otherwise be ambiguous.

Example

#include <iostream>
#include <chrono>
 
int main()
{
    std::cout << std::boolalpha;
 
    auto ymdl {11/std::chrono::last/2020};
    ymdl = std::chrono::years(10) + ymdl;
    std::cout << (ymdl == std::chrono::day(30)/
                          std::chrono::November/
                          std::chrono::year(2030)) << ' ';
    ymdl = ymdl - std::chrono::months(6);
    std::cout << (ymdl == std::chrono::day(31)/
                          std::chrono::May/
                          std::chrono::year(2030)) << '\n';
}

Output:

true true