operator+,-,*,/,%(std::chrono::duration)
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template<class Rep1, class Period1, class Rep2, class Period2 > typenamestd::common_type<duration<Rep1,Period1>, duration<Rep2,Period2>>::type | (1) | |
template<class Rep1, class Period1, class Rep2, class Period2 > typenamestd::common_type<duration<Rep1,Period1>, duration<Rep2,Period2>>::type | (2) | |
template<class Rep1, class Period, class Rep2 > duration<typenamestd::common_type<Rep1,Rep2>::type, Period> | (3) | |
template<class Rep1, class Rep2, class Period > duration<typenamestd::common_type<Rep1,Rep2>::type, Period> | (4) | |
template<class Rep1, class Period, class Rep2 > duration<typename common_type<Rep1,Rep2>::type, Period> | (5) | |
template<class Rep1, class Period1, class Rep2, class Period2 > typenamestd::common_type<Rep1,Rep2>::type | (6) | |
template<class Rep1, class Period, class Rep2 > duration<typename common_type<Rep1,Rep2>::type, Period> | (7) | |
template<class Rep1, class Period1, class Rep2, class Period2 > typename common_type<duration<Rep1,Period1>, duration<Rep2,Period2>>::type | (8) | |
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rhs
Anzahl der Ticks aus der lhs
Anzahl der Ticks nach der Konvertierung abgezogen .rhs
number of ticks subtracted from the lhs
number of ticks after conversion.You can help to correct and verify the translation. Click here for instructions.
d
um ein, deren rep
ist die häufigste Art von Rep1
und Rep2
und Multiples die Anzahl der Ticks nach Umsetzung durch s
.d
to one whose rep
is the common type between Rep1
and Rep2
, and multiples the number of ticks after conversion by s
.You can help to correct and verify the translation. Click here for instructions.
d
um ein, deren rep
ist die häufigste Art von Rep1
und Rep2
und teilt die Anzahl der Ticks nach Umsetzung durch s
d
to one whose rep
is the common type between Rep1
and Rep2
, and divides the number of ticks after conversion by s
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lhs
nach der Konvertierung durch die Taktanzahl der rhs
nach der Konvertierung. Beachten Sie, dass der Rückgabewert dieses Operators nicht eine Dauer .lhs
after conversion by the tick count of rhs
after conversion. Note that the return value of this operator is not a duration.You can help to correct and verify the translation. Click here for instructions.
d
um ein, deren rep
ist die häufigste Art von Rep1
und Rep2
und schafft eine Laufzeit, deren Taktanzahl ist der Rest der Division der Anzahl der Ticks, nach der Konvertierung durch s
.d
to one whose rep
is the common type between Rep1
and Rep2
, and creates a duration whose tick count is the remainder of the division of the tick count, after conversion, by s
.You can help to correct and verify the translation. Click here for instructions.
You can help to correct and verify the translation. Click here for instructions.
[Bearbeiten]Parameter
lhs | - | Dauer auf der linken Seite des Fahrers Original: duration on the left-hand side of the operator The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
rhs | - | Dauer auf der rechten Seite des Fahrers Original: duration on the right-hand side of the operator The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
d | - | die Dauer Argument für Mixed-Argument Betreiber Original: the duration argument for mixed-argument operators The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
s | - | Taktanzahl Argument für Mixed-Argument Betreiber Original: tick count argument for mixed-argument operators The text has been machine-translated via Google Translate. You can help to correct and verify the translation. Click here for instructions. |
[Bearbeiten]Rückgabewert
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[Bearbeiten]Beispiel
#include <chrono>#include <iostream>int main(){// simple arithmeticstd::chrono::seconds s =std::chrono::hours(1)+2*std::chrono::minutes(10)+std::chrono::seconds(70)/10; std::cout<<"1 hour + 2*10 min + 70/10 sec = "<< s.count()<<" seconds\n"; // difference between dividing a duration by a number// and dividing a duration by another durationstd::cout<<"Dividing that by 2 minutes gives "<< s /std::chrono::minutes(2)<<'\n';std::cout<<"Dividing that by 2 gives "<<(s /2).count()<<" sconds\n"; // the remainder operator is useful in determining where in a time// frame is this particular duration, e.g. to break it down into hours,// minutes, and seconds:std::cout<< s.count()<<" seconds is "<<std::chrono::duration_cast<std::chrono::hours>( s ).count()<<" hours, "<<std::chrono::duration_cast<std::chrono::minutes>( s %std::chrono::hours(1)).count()<<" minutes, "<<std::chrono::duration_cast<std::chrono::seconds>( s %std::chrono::minutes(1)).count()<<" seconds\n";}
Output:
1 hour + 2*10 min + 70/10 sec = 4807 seconds Dividing that by 2 minutes gives 40 Dividing that by 2 gives 2403 sconds 4807 seconds is 1 hours, 20 minutes, 7 seconds