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std::chrono::year_month_day::operator+=, std::chrono::year_month_day::operator-=

From cppreference.com
 
 
Utilities library
General utilities
Relational operators (deprecated in C++20)
 
 
 
constexpr std::chrono::year_month_day&
    operator+=( const std::chrono::years& dy ) const noexcept;
(1) (since C++20)
constexpr std::chrono::year_month_day&
    operator+=( const std::chrono::months& dm ) const noexcept;
(2) (since C++20)
constexpr std::chrono::year_month_day&
    operator-=( const std::chrono::years& dy ) const noexcept;
(3) (since C++20)
constexpr std::chrono::year_month_day&
    operator-=( const std::chrono::months& dm ) const noexcept;
(4) (since C++20)

Modifies the time point *this represents by the duration dy or dm.

1) Equivalent to *this = *this + dy;.
2) Equivalent to *this = *this + dm;.
3) Equivalent to *this = *this - dy;.
4) Equivalent to *this = *this - dm;.

For durations that are convertible to both std::chrono::years and std::chrono::months, the years overloads (1,3) are preferred if the call would otherwise be ambiguous.

[edit] Example

#include <cassert>
#include <chrono>
#include <iostream>
 
int main()
{
    constexpr auto monthsInYear{12};
    auto ymd{std::chrono::day(1)/std::chrono::July/2020};
    std::cout << "#1 " << ymd << '\n';
 
    ymd -= std::chrono::years(10);
    std::cout << "#2 " << ymd << '\n';
    assert(ymd.month() == std::chrono::July);
    assert(ymd.year() == std::chrono::year(2010));
 
    ymd += std::chrono::months(10 * monthsInYear + 11);
    std::cout << "#3 " << ymd << '\n';
    assert(ymd.month() == std::chrono::month(6));
    assert(ymd.year() == std::chrono::year(2021));
 
    // Handling the ymd += months "overflow" case.
    ymd = std::chrono::May/31/2021; // ok
    std::cout << "#4 " << ymd << '\n';
    assert(ymd.ok());
 
    ymd += std::chrono::months{1}; // bad date: June has only 30 days
    std::cout << "#5 " << ymd << '\n';
    assert(not ymd.ok());
    assert(ymd == std::chrono::June/31/2021);
 
    // Snap to the last day of the month, June 30:
    const auto ymd1 = ymd.year()/ymd.month()/std::chrono::last;
    std::cout << "#6 " << ymd1 << '\n';
    assert(ymd1.ok());
    assert(ymd1 == std::chrono::June/30/2021);
 
    // Overflow into the next month, July 1 (via converting to/from sys_days):
    const std::chrono::year_month_day ymd2 = std::chrono::sys_days{ymd};
    std::cout << "#7 " << ymd2 << '\n';
    assert(ymd2.ok());
    assert(ymd2 == std::chrono::July/1/2021);
}

Output:

#1 2020-07-01
#2 2010-07-01
#3 2021-06-01
#4 2021-05-31
#5 2021-06-31 is not a valid date
#6 2021/Jun/last
#7 2021-07-01

[edit] See also

adds or subtracts a year_month_day and some number of years or months
(function) [edit]