std::chrono::year_month_day_last:: operator sys_days, std::chrono::year_month_day_last:: operator local_days
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cpp
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chrono
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year month day last
C++
Date and time library
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std::chrono::year_month_day_last
| Member functions | ||||
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year_month_day_last::operator sys_days
year_month_day_last::operator local_days
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| Nonmember functions | ||||
| Helper classes | ||||
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constexpr
operator
std::
chrono
::
sys_days
(
)
const
noexcept
;
|
(1) | (seit C++20) |
|
constexpr
explicit
operator
std::
chrono
::
local_days
(
)
const
noexcept
;
|
(2) | (seit C++20) |
Konvertiert
*
this
zu einem
std::chrono::time_point
, der dasselbe Datum wie dieses
year_month_day_last
repräsentiert. Dies entspricht dem Zusammensetzen eines
year_month_day
aus
year()
,
month()
und
day()
und der Konvertierung dieses
year_month_day
in den Zieltyp.
1)
Entspricht
std::
chrono
::
sys_days
(
year
(
)
/
month
(
)
/
day
(
)
)
.
2)
Entspricht
std::
chrono
::
local_days
(
year
(
)
/
month
(
)
/
day
(
)
)
.
Beispiel
Diesen Code ausführen
#include <array> #include <chrono> #include <iostream> #include <string_view> using namespace std::chrono; using namespace std::literals; int main() { constexpr std::chrono::year y{2023y}; constexpr std::array quarters{"1st"sv, "2nd"sv, "3rd"sv, "4th"sv}; constexpr auto mq{12 / 4}; // months per quarter std::cout << "In year " << static_cast<int>(y) << '\n'; for (auto q = 1; q < 5; ++q) { const auto ls = y / std::chrono::month(q * mq) / Sunday[last]; const auto ld = y / std::chrono::month(q * mq) / last; // subtract last Sunday from last day for day of week const auto index = (sys_days(ld) - sys_days(ls)).count(); std::cout << "The " << quarters[q - 1] << " quarter ends on a " << std::chrono::weekday(index) << '\n'; } }
Ausgabe:
In year 2023 The 1st quarter ends on a Fri The 2nd quarter ends on a Fri The 3rd quarter ends on a Sat The 4th quarter ends on a Sun
Siehe auch
|
Konvertiert zu einem
std::chrono::time_point
(öffentliche Elementfunktion von
std::chrono::year_month_day
)
|