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

    From zh.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)