std::chrono::year_month_weekday_last::operator+=, std::chrono::year_month_weekday_last::operator-=
From cppreference.com
< cpp | chrono | year month weekday last
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
.
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
(C++20) |
adds or subtracts a year_month_weekday_last and some number of years or months (function) |