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std::chrono::year_month_weekday_last::operator+=,(3) C++ Standard Libary std::chrono::year_month_weekday_last::operator+=,(3)

NAME

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

Synopsis


constexpr std::chrono::year_month_weekday_last& (1) (since C++20)
operator+=( const std::chrono::years& dy ) const noexcept;
constexpr std::chrono::year_month_weekday_last& (2) (since C++20)
operator+=( const std::chrono::months& dm ) const noexcept;
constexpr std::chrono::year_month_weekday_last& (3) (since C++20)
operator-=( const std::chrono::years& dy ) const noexcept;
constexpr std::chrono::year_month_weekday_last& (4) (since C++20)
operator-=( const std::chrono::months& dm ) const noexcept;


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.

Example

// Run this code


#include <iostream>
#include <chrono>
using namespace std::chrono;


int main()
{
std::cout << std::boolalpha;


auto ymwdl {August/Tuesday[last]/2022};
ymwdl += months(2);
std::cout << (year_month_day{ymwdl} == October/25/2022) << ' ';
ymwdl -= years(1);
std::cout << (year_month_day{ymwdl} == October/26/2021) << '\n';
}

Output:


true true

See also


operator+ adds or subtracts a year_month_weekday_last and some number of years or
operator- months
(C++20) (function)

2022.07.31 http://cppreference.com