constexpr std::chrono::year_month_weekday_last& operator+=( const std::chrono::years& dy ) const noexcept; | (1) | (since C++20) |
constexpr std::chrono::year_month_weekday_last& operator+=( const std::chrono::months& dm ) const noexcept; | (2) | (since C++20) |
constexpr std::chrono::year_month_weekday_last& operator-=( const std::chrono::years& dy ) const noexcept; | (3) | (since C++20) |
constexpr std::chrono::year_month_weekday_last& operator-=( const std::chrono::months& dm ) const noexcept; | (4) | (since C++20) |
Modifies the time point *this
represents by the duration dy
or dm
.
*this = *this + dy;
*this = *this + dm;
*this = *this - dy;
*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.
#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
(C++20) | adds or subtracts a year_month_weekday_last and some number of years or months (function) |
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