std::chrono::operator<<(std::chrono::day) (3) Linux Manual Page

std::chrono::operator==,!=,<,<=,>,>=(std::chrono::year_month_day_last) – std::chrono::operator==,!=,<,<=,>,>=(std::chrono::year_month_day_last)

Synopsis

constexpr bool operator==(const std::chrono::year_month_day_last &x, (1)(since C++ 20)

                                                                         const std::chrono::year_month_day_last &y) noexcept;

constexpr bool operator!=(const std::chrono::year_month_day_last &x, (2)(since C++ 20)

                                                                         const std::chrono::year_month_day_last &y) noexcept;

constexpr bool operator<(const std::chrono::year_month_day_last &x, (3)(since C++ 20)

                                                                        const std::chrono::year_month_day_last &y) noexcept;

constexpr bool operator>(const std::chrono::year_month_day_last &x, (4)(since C++ 20)

                                                                        const std::chrono::year_month_day_last &y) noexcept;

constexpr bool operator<=(const std::chrono::year_month_day_last &x, (5)(since C++ 20)

                                                                         const std::chrono::year_month_day_last &y) noexcept;

constexpr bool operator>=(const std::chrono::year_month_day_last &x, (6)(since C++ 20)

                                                                         const std::chrono::year_month_day_last &y) noexcept;

Compares the two year_month_day_last objects x and y. This is a lexicographical comparison: the year() is compared first, then month().

Return value

1) x.year() == y.year() && x.month() == y.month()
2) !(x == y)
3) If x.year() != y.year(), x.year() < y.year(); otherwise, x.month() < y.month().
4) y < x
5) !(y < x)
6) !(x < y)

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.