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

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Duration
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Calendar
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chrono I/O
(C++20)
C-style date and time
std::chrono::year_month_day_last
Defined in header <chrono>
(since C++20)
(since C++20)
(since C++20)
(since C++20)
(since C++20)
(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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