# std::chrono::operator+, std::chrono::operator- (std::chrono::year)

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< cpp‎ | chrono‎ | year

constexpr std::chrono::year operator+(const std::chrono::year & y, const std::chrono::years & ys) noexcept; | (1) | (since C++20)
constexpr std::chrono::year operator+(const std::chrono::years & ys, const std::chrono::year & y) noexcept; | (2) | (since C++20)
constexpr std::chrono::year operator-(const std::chrono::year & y, const std::chrono::years & ys) noexcept; | (3) | (since C++20)
constexpr std::chrono::years operator-(const std::chrono::year & y1, const std::chrono::year & y2) noexcept; | (4) | (since C++20)

1-2) Adds `ys.count()` years to `y`.
3) Subtracts `ys.count()` years from `y`.
4) Returns the difference in years between `y1` and `y2`.

### Return value

1-2) std::chrono::year (int(y) + ys.count())
3) std::chrono::year (int(y) - ys.count())
4) std::chrono::years (int(y1) - int(y2))

### Notes

If the resulting year value for (1-3) is outside the range [-32767,32767], the actual value stored is unspecified.

The result of subtracting two `year` values is a duration of type std::chrono::years. This duration unit represents the length of the average Gregorian year, and the resulting duration bears no relationship to the number of days in the particular years represented by the operands. For example, the result of 2018y - 2017y is std::chrono::years (1), which represents 365.2425 days, not 365 days.

### Example

| This section is incomplete Reason: no example

### See also

operator++operator++(int)operator--operator--(int) | increments or decrements the month (public member function of `std::chrono::month`)
operator+=operator-= | adds or subtracts a number of months (public member function of `std::chrono::month`)

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