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

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< cpp‎ | chrono‎ | year month day last

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

Run this code

    #include <cassert>
    #include <chrono>
    #include <iostream>

    int main()
    {
        auto ymdl{11/std::chrono::last/2020};
        std::cout << ymdl << '\n';

        ymdl = std::chrono::years(10) + ymdl;
        std::cout << ymdl << '\n';
        assert(ymdl == std::chrono::day(30)/
                       std::chrono::November/
                       std::chrono::year(2030));

        ymdl = ymdl - std::chrono::months(6);
        std::cout << ymdl << '\n';
        assert(ymdl == std::chrono::day(31)/
                       std::chrono::May/
                       std::chrono::year(2030));
    }

Output:

    2020/Nov/last
    2030/Nov/last
    2030/May/last

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