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

From cppreference.net

C++
Date and time library
std::chrono::year_month_weekday
constexpr std:: chrono :: year_month_weekday &
operator + = ( const std:: chrono :: years & dy ) const noexcept ;
(1) (自 C++20 起)
constexpr std:: chrono :: year_month_weekday &
operator + = ( const std:: chrono :: months & dm ) const noexcept ;
(2) (自 C++20 起)
constexpr std:: chrono :: year_month_weekday &
operator - = ( const std:: chrono :: years & dy ) const noexcept ;
(3) (自 C++20 起)
constexpr std:: chrono :: year_month_weekday &
operator - = ( const std:: chrono :: months & dm ) const noexcept ;
(4) (自 C++20 起)

修改 * this 所表示的时间点，增减时长为 dy 或 dm 。

1) 等价于 * this = * this + dy ; 。
2) 等价于 * this = * this + dm ; 。
3) 等价于 * this = * this - dy ; 。
4) 等价于 * this = * this - dm ; 。

对于可同时转换为 std::chrono::years 和 std::chrono::months 的时间段，若调用存在歧义，则优先选择 years 重载 (1,3) 。

示例

运行此代码 
#include <cassert>
#include <chrono>
#include <iostream>
int main()
{
    auto ymwi{1/std::chrono::Wednesday[2]/2021};
    std::cout << ymwi << '\n';
    ymwi += std::chrono::years(5);
    std::cout << ymwi << '\n';
    assert(static_cast<std::chrono::year_month_day>(ymwi) ==
                       std::chrono::year(2026)/1/14);
    ymwi -= std::chrono::months(1);
    std::cout << ymwi << '\n';
    assert(static_cast<std::chrono::year_month_day>(ymwi) == 
                       std::chrono::day(10)/12/2025);
}

输出： 

2021/Jan/Wed[2]
2026/Jan/Wed[2]
2025/Dec/Wed[2]

参见

(C++20)
对 year_month_weekday 进行加减若干年或月的运算
(函数)