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

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std::chrono::year_month_day
 
Defined in header <chrono>
constexpr std::chrono::year_month_day operator+( const std::chrono::year_month_day& ymd,

                                                 const std::chrono::months& dm

                                               ) noexcept;
 (since C++20)
constexpr std::chrono::year_month_day operator+( const std::chrono::months& dm,

                                                 const std::chrono::year_month_day& ymd

                                               ) noexcept;
 (since C++20)
constexpr std::chrono::year_month_day operator+( const std::chrono::year_month_day& ymd,

                                                 const std::chrono::years& dy

                                               ) noexcept;
 (since C++20)
constexpr std::chrono::year_month_day operator+( const std::chrono::years& dy,

                                                 const std::chrono::year_month_day& ymd

                                               ) noexcept;
 (since C++20)
constexpr std::chrono::year_month_day operator-( const std::chrono::year_month_day& ymd,

                                                 const std::chrono::months& dm

                                               ) noexcept;
 (since C++20)
constexpr std::chrono::year_month_day operator-( const std::chrono::year_month_day& ymd,

                                                 const std::chrono::years& dy

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

Notes

Even if ymd.ok() is true, the resulting year_month_day may not represent a valid date if ymd.day() is 29, 30, or 31.

Example

Run this code
#include <iostream>
#include <chrono>
 
int main()
{
    std::cout << std::boolalpha;
 
    auto ymd {std::chrono::day(1)/std::chrono::July/2021};
 
    ymd = ymd + std::chrono::months(4);
    std::cout << (ymd.month() == std::chrono::November) << ' '
              << (ymd.year() == std::chrono::year(2021)) << ' ';
 
    ymd = ymd - std::chrono::years(10);
    std::cout << (ymd.month() == std::chrono::month(11)) << ' '
              << (ymd.year() == std::chrono::year(2011)) << ' ';
}

Output: 

true true true true
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