std::chrono::operator+, std::chrono::operator- (std::chrono::weekday)
From cppreference.com
constexpr std::chrono::weekday operator+(const std::chrono::weekday& wd, const std::chrono::days& d) noexcept; |
(1) | (since C++20) |
constexpr std::chrono::weekday operator+(const std::chrono::days& d, const std::chrono::weekday& wd) noexcept; |
(2) | (since C++20) |
constexpr std::chrono::weekday operator-(const std::chrono::weekday& wd, const std::chrono::days& d) noexcept; |
(3) | (since C++20) |
constexpr std::chrono::days operator-(const std::chrono::weekday& wd1, const std::chrono::weekday& wd2) noexcept; |
(4) | (since C++20) |
1-2) Adds
d.count()
days to wd
. The weekday value held in the result is computed by first evaluating static_cast<long long>(unsigned(wd)) + d.count() and reducing it modulo 7 to an integer in the range [0, 6].3) Subtracts
d.count()
days from wd
. Equivalent to return wd + -d;4) If wd1.ok() and wd2.ok() are both true, returns a std::chrono::days value
d
such that d.count()
is in the range [0, 6] and wd2 + d == wd1. Otherwise the returned value is unspecified.Return value
1-3) A std::chrono::weekday holding a weekday value calculated as described above.
4) A std::chrono::days representing the distance between
wd1
and wd2
.Notes
As long as the computation doesn't overflow, (1-3) always return a valid weekday
even if wd.ok() is false.
Example
This section is incomplete Reason: no example |
See also
increments or decrements the weekday (public member function) | |
adds or subtracts a number of days (public member function) |