std::chrono::operator+, std::chrono::operator- (std::chrono::year_month_day_last)
From cppreference.com
< cpp | chrono | year month day last
Defined in header <chrono>
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constexpr std::chrono::year_month_day_last operator+( const std::chrono::year_month_day_last& ymdl, const std::chrono::months& dm ) |
(since C++20) | |
constexpr std::chrono::year_month_day_last operator+( const std::chrono::months& dm, const std::chrono::year_month_day_last& ymdl ) |
(since C++20) | |
constexpr std::chrono::year_month_day_last operator+( const std::chrono::year_month_day_last& ymdl, const std::chrono::years& dy ) |
(since C++20) | |
constexpr std::chrono::year_month_day_last operator+( const std::chrono::years& dy, const std::chrono::year_month_day_last& ymdl ) |
(since C++20) | |
constexpr std::chrono::year_month_day_last operator-( const std::chrono::year_month_day_last& ymdl, const std::chrono::months& dm ) |
(since C++20) | |
constexpr std::chrono::year_month_day_last operator-( const std::chrono::year_month_day_last& ymdl, const std::chrono::years& dy ) |
(since C++20) | |
1-2) Adds dm.count() months to the date represented by
ymdl
. The result has the same year()
and month()
as std::chrono::year_month(ymdl.year(), ymdl.month()) + dm.3-4) Adds dy.count() years to the date represented by
ymdl
. The result is equivalent to std::chrono::year_month_day_last(ymdl.year() + dy, ymdl.month_day_last()).5) Subtracts dm.count() months from the date represented by
ymdl
. Equivalent to ymdl + -dm.6) Subtracts dy.count() years from the date represented by
ymdl
. Equivalent to ymdl + -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> int main() { std::cout << std::boolalpha; auto ymdl {11/std::chrono::last/2020}; ymdl = std::chrono::years(10) + ymdl; std::cout << (ymdl == std::chrono::day(30)/ std::chrono::November/ std::chrono::year(2030)) << ' '; ymdl = ymdl - std::chrono::months(6); std::cout << (ymdl == std::chrono::day(31)/ std::chrono::May/ std::chrono::year(2030)) << '\n'; }
Output:
true true