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.
See also
adds or subtracts a year_month_day and some number of years or months (public member function) |
Example
Run this code
#include <iostream> #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) << ' ' << (ymd.year() == std::chrono::year(2010)) << ' '; ymd += std::chrono::months(10 * monthsInYear + 11); std::cout << (ymd.month() == std::chrono::month(6)) << ' ' << (ymd.year() == std::chrono::year(2021)) << ' '; }
Output:
true true true true