std::chrono::year_month_day::operator+=, std::chrono::year_month_day::operator-=
From cppreference.com
< cpp | chrono | year month day
constexpr std::chrono::year_month_day& operator+=( const std::chrono::years& dy ) const noexcept; |
(1) | (since C++20) |
constexpr std::chrono::year_month_day& operator+=( const std::chrono::months& dm ) const noexcept; |
(2) | (since C++20) |
constexpr std::chrono::year_month_day& operator-=( const std::chrono::years& dy ) const noexcept; |
(3) | (since C++20) |
constexpr std::chrono::year_month_day& 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 <cassert> #include <chrono> int main() { std::cout << std::boolalpha; constexpr auto monthsInYear {12}; auto ymd {std::chrono::day(1)/std::chrono::July/2020}; ymd -= std::chrono::years(10); std::cout << (ymd.month() == std::chrono::July) << ' ' // true << (ymd.year() == std::chrono::year(2010)) << ' '; // true ymd += std::chrono::months(10 * monthsInYear + 11); std::cout << (ymd.month() == std::chrono::month(6)) << ' ' // true << (ymd.year() == std::chrono::year(2021)) << ' '; // true // Handling the ymd += months "overflow" case. ymd = std::chrono::May/31/2021; // ok std::cout << ymd.ok() << ' '; // true ymd += std::chrono::months{1}; // bad date: June has only 30 days std::cout << ymd.ok() << ' '; // false assert(ymd == std::chrono::June/31/2021); // Snap to the last day of the month, June 30: const auto ymd1 = ymd.year()/ymd.month()/std::chrono::last; std::cout << ymd1.ok() << ' '; // true assert(ymd1 == std::chrono::June/30/2021); // Overflow into the next month, July 1 (via converting to/from sys_days): const std::chrono::year_month_day ymd2 = std::chrono::sys_days{ymd}; std::cout << ymd2.ok() << ' '; // true assert(ymd2 == std::chrono::July/1/2021); }
Output:
true true true true true false true true
See also
(C++20) |
adds or subtracts a year_month_day and some number of years or months (function) |