std::chrono::operator+, std::chrono::operator- (std::chrono::year_month_weekday_last)
From cppreference.com
< cpp | chrono | year month weekday last
constexpr std::chrono::year_month_weekday_last operator+( const std::chrono::year_month_weekday_last& ymwdl, |
(since C++20) | |
constexpr std::chrono::year_month_weekday_last operator+( const std::chrono::months& dm, |
(since C++20) | |
constexpr std::chrono::year_month_weekday_last operator+( const std::chrono::year_month_weekday_last& ymwdl, |
(since C++20) | |
constexpr std::chrono::year_month_weekday_last operator+( const std::chrono::years& dy, |
(since C++20) | |
constexpr std::chrono::year_month_weekday_last operator-( const std::chrono::year_month_weekday_last& ymwdl, |
(since C++20) | |
constexpr std::chrono::year_month_weekday_last operator-( const std::chrono::year_month_weekday_last& ymwdl, |
(since C++20) | |
1,2) Adds dm.count() months to the date represented by ymwdl. The result has the same
year()
and month()
as std::chrono::year_month(ymwdl.year(), ymwdl.month()) + dm and the same weekday()
as ymwdl.3,4) Adds dy.count() years to the date represented by ymwdl. The result is equivalent to std::chrono::year_month_weekday_last(ymwdl.year() + dy, ymwdl.month(), ymwd.weekday_last()).
5) Subtracts dm.count() months from the date represented by ymwdl. Equivalent to ymwdl + -dm.
6) Subtracts dy.count() years from the date represented by ymwdl. Equivalent to ymwdl + -dy.
For durations that are convertible to both std::chrono::years and std::chrono::months, the years
overloads (3,4,6) are preferred if the call would otherwise be ambiguous.
Example
Run this code
#include <cassert> #include <chrono> using namespace std::chrono; int main() { constexpr auto ymwdl1{Tuesday[last]/11/2021}; auto ymwdl2 = ymwdl1; ymwdl2 = std::chrono::months(12) + ymwdl2; ymwdl2 = ymwdl2 - std::chrono::years(1); assert(ymwdl1 == ymwdl2); }