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 <iostream> #include <chrono> using namespace std::chrono; int main() { std::cout << std::boolalpha; constexpr auto ymwdl1 {Tuesday[last]/11/2021}; auto ymwdl2 = ymwdl1; ymwdl2 = months(12) + ymwdl2; ymwdl2 = ymwdl2 - years(1); std::cout << (ymwdl1 == ymwdl2) << '\n'; }
Output:
true