std::chrono::operator==,<=>(std::chrono::year_month_day)
From cppreference.com
< cpp | chrono | year month day
Defined in header <chrono>
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constexpr bool operator==( const std::chrono::year_month_day& x, const std::chrono::year_month_day& y ) noexcept; |
(1) | (since C++20) |
constexpr std::strong_ordering operator<=>( const std::chrono::year_month_day& x, |
(2) | (since C++20) |
Compares the two year_month_day
values x and y. This is a lexicographical comparison: the year()
is compared first, then month()
, then day()
.
The <
, <=
, >
, >=
, and !=
operators are synthesized from operator<=> and operator== respectively.
[edit] Return value
1) x.year() == y.year() && x.month() == y.month() && x.day() == y.day()
2) If x.year() <=> y.year != 0, x.year() <=> y.year; otherwise if x.month() <=> y.month() != 0, x.month() <=> y.month(); otherwise x.day() <=> y.day().
[edit] Notes
If both x and y represent valid dates (x.ok() && y.ok() == true), the result of the lexicographical comparison is consistent with the calendar order.
[edit] Example
Run this code
#include <chrono> int main() { constexpr auto ymd1{std::chrono::day(13)/7/1337}; constexpr auto ymd2{std::chrono::year(1337)/7/13}; static_assert(ymd1 == ymd2); static_assert(ymd1 <= ymd2); static_assert(ymd1 >= ymd2); static_assert(ymd1 <=> ymd2 == 0); }