std::chrono::operator+, std::chrono::operator- (std::chrono::year_month)
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< cpp | chrono | year month
constexpr std::chrono::year_month operator+(const std::chrono::year_month& ym, const std::chrono::years& dy) noexcept; |
(1) | (since C++20) |
constexpr std::chrono::year_month operator+(const std::chrono::years& dy, const std::chrono::year_month& ym) noexcept; |
(2) | (since C++20) |
constexpr std::chrono::year_month operator+(const std::chrono::year_month& ym, const std::chrono::months& dm) noexcept; |
(3) | (since C++20) |
constexpr std::chrono::year_month operator+(const std::chrono::months& dm, const std::chrono::year_month& ym) noexcept; |
(4) | (since C++20) |
constexpr std::chrono::year_month operator-(const std::chrono::year_month& ym, const std::chrono::years& dy) noexcept; |
(5) | (since C++20) |
constexpr std::chrono::year_month operator-(const std::chrono::year_month& ym, const std::chrono::months& dm) noexcept; |
(6) | (since C++20) |
constexpr std::chrono::months operator-(const std::chrono::year_month& ym1, const std::chrono::year_month& ym2) noexcept; |
(7) | (since C++20) |
1-2) Adds
dy.count()
years to ym
. 3-4) Adds
dm.count()
months to ym
. 5) Subtracts
dy.count()
years from ym
.6) Subtracts
dm.count()
months from ym
.7) Returns the difference in months between the two time points represented by
ym1
and ym2
.For durations that are convertible to both std::chrono::years and std::chrono::months, the years
overloads (1,2,5) are preferred if the call would otherwise be ambiguous.
Return value
1-2) std::chrono::year_month(ym.year() + dy, ym.month())
3-4) A
year_month
value z
such that z - ym == dm and z.ok() == true.5) ym + -dy
6) ym + -dm
7) ym1.year() - ym2.year() + std::chrono::months(int(unsigned(ym1.month())) - int(unsigned(ym2.month())))
Notes
The result of subtracting two year_month
values is a duration of type std::chrono::months. This duration unit represents the length of the average Gregorian month (30.436875 days), and the resulting duration bears no relationship to the actual number of days in the time period at issue. For example, the result of 2017y/3 - 2017y/2 is std::chrono::months(1), even though February 2017 only contains 28 days.
Example
Run this code
#include <iostream> #include <chrono> int main() { std::cout << std::boolalpha; auto ym {std::chrono::year(2021)/std::chrono::July}; ym = ym + std::chrono::months(14); std::cout << (ym.month() == std::chrono::September) << ' ' << (ym.year() == std::chrono::year(2022)) << ' '; ym = ym - std::chrono::years(3); std::cout << (ym.month() == std::chrono::month(9)) << ' ' << (ym.year() == std::chrono::year(2019)) << ' '; ym = ym + (std::chrono::September - std::chrono::month(2)); std::cout << (ym.month() == std::chrono::April) << ' ' << (ym.year() == std::chrono::year(2020)) << '\n'; }
Output:
true true true true true true
See also
modifies the year_month by some number of months or years (public member function) |