NAME¶

std::chrono::operator+,std::chrono::operator-(std::chrono::year_month) - std::chrono::operator+,std::chrono::operator-(std::chrono::year_month)

Synopsis¶

constexpr std::chrono::year_month operator+(const
std::chrono::year_month& ym, (1) (since C++20)
const std::chrono::years& dy) noexcept;
constexpr std::chrono::year_month operator+(const
std::chrono::years& dy, (2) (since C++20)
const std::chrono::year_month& ym) noexcept;
constexpr std::chrono::year_month operator+(const
std::chrono::year_month& ym, (3) (since C++20)
const std::chrono::months& dm) noexcept;
constexpr std::chrono::year_month operator+(const
std::chrono::months& dm, (4) (since C++20)
const std::chrono::year_month& ym) noexcept;
constexpr std::chrono::year_month operator-(const
std::chrono::year_month& ym, (5) (since C++20)
const std::chrono::years& dy) noexcept;
constexpr std::chrono::year_month operator-(const
std::chrono::year_month& ym, (6) (since C++20)
const std::chrono::months& dm) noexcept;
constexpr std::chrono::months operator-(const
std::chrono::year_month& ym1, (7) (since C++20)
const std::chrono::year_month& ym2) noexcept;

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¶

operator+= modifies the year_month by some number of months or years
operator-= (public member function)