Time Duration

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A time duration is the difference between two time points. It will be measured in time ticks.

Time duration

Time duration is a class template. std::chrono::duration consists of the type of the tick Rep and the length of a tick Period.

template<
    class Rep,
    class Period = std::ratio<1>
> class duration;

 

The tick length is by default std::ratio<1> std::ratio<1> stands for a second and can also be written as std::ratio<1,1>. Therefore, it's quite easy. std::ratio<60> is a minute and std::ratio<1,1000> a millisecond. When Rep is a floating point number, you can use it to hold fractions of time ticks.

C++11 predefines the most important time durations:

typedef duration<signed int, nano> nanoseconds;
typedef duration<signed int, micro> microseconds;
typedef duration<signed int, milli> milliseconds;
typedef duration<signed int> seconds;
typedef duration<signed int, ratio< 60>> minutes;
typedef duration<signed int, ratio<3600>> hours;

 

How much time has passed since the UNIX epoch (1.1.1970)? Thanks to type aliases for the different time durations I can answer the question quite easily. For simplicity reasons I ignore leap years and assume that a year has 365 days.

 

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// timeSinceEpoch.cpp

#include <chrono>
#include <iostream>

int main(){

  std::cout << std::fixed << std::endl;
  
  std::cout << "Time since 1.1.1970:\n" << std::endl;

  auto timeNow= std::chrono::system_clock::now();
  auto duration= timeNow.time_since_epoch();
  std::cout << duration.count() << " nanoseconds " << std::endl;

  typedef std::chrono::duration<long double,std::ratio<1,1000000>> MyMicroSecondTick;
  MyMicroSecondTick micro(duration);
  std::cout << micro.count() << " microseconds" << std::endl;
  
  typedef std::chrono::duration<long double,std::ratio<1,1000>> MyMilliSecondTick;
  MyMilliSecondTick milli(duration);
  std::cout << milli.count() << " milliseconds" << std::endl;
  
  typedef std::chrono::duration<long double> MySecondTick;
  MySecondTick sec(duration);
  std::cout << sec.count() << " seconds " << std::endl;
  
  typedef std::chrono::duration<double, std::ratio<60>> MyMinuteTick;
  MyMinuteTick myMinute(duration);
  std::cout << myMinute.count() << " minutes" << std::endl;

  typedef std::chrono::duration<double, std::ratio<60*60>> MyHourTick;
  MyHourTick myHour(duration);
  std::cout << myHour.count() << " hours" << std::endl;
  
  typedef std::chrono::duration<double, std::ratio<60*60*24*365>> MyYearTick;
  MyYearTick myYear(duration);
  std::cout << myYear.count() << " years" << std::endl;

  typedef std::chrono::duration<double, std::ratio<60*45>> MyLessonTick;
  MyLessonTick myLesson(duration);
  std::cout << myLesson.count() << " lessons" << std::endl;

  std::cout << std::endl;

}

 

In addition to the typical time durations microsecond (line 16), millisecond (line 20), second (line 24), minute (line 28), hour (line 32), and year (line 36) I define the German school hour (45 min) in line 45.

 timeSinceEpoch

It's quite convenient to calculate with time durations.

Calculations with time durations

The time durations support basic mathematics. The means in the case of multiplication or division that you can multiply or divide a time duration by a number. Of course, you can compare time durations. I explicitly want to emphasise: all these calculations and comparisons respect the units of the time durations.

It gets even better with C++14 because C++14 has a bunch of predefined time literals.

timeLiteralsEng

How much time needs my 16 years old son for his typical school day? I will answer the question in the example and show the result in different time durations.

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// schoolDay.cpp

#include <iostream>
#include <chrono>

using namespace std::literals::chrono_literals;

int main(){

  std::cout << std::endl;

  constexpr auto schoolHour= 45min;

  constexpr auto shortBreak= 300s;
  constexpr auto longBreak= 0.25h;

  constexpr auto schoolWay= 15min;
  constexpr auto homework= 2h;

  constexpr auto schoolDayInSeconds= 2*schoolWay + 6 * schoolHour + 4 * shortBreak + longBreak + homework;

  std::cout << "School day in seconds: " << schoolDayInSeconds.count() << std::endl;

  std::chrono::duration<double,std::ratio<3600>> schoolDayInHours = schoolDayInSeconds;
  std::chrono::duration<double,std::ratio<60>> schoolDayInMinutes = schoolDayInSeconds;
  std::chrono::duration<double,std::ratio<1,1000>> schoolDayInMilliseconds= schoolDayInSeconds;

  std::cout << "School day in hours: " << schoolDayInHours.count() << std::endl;
  std::cout << "School day in minutes: " << schoolDayInMinutes.count() << std::endl;
  std::cout << "School day in milliseconds: " << schoolDayInMilliseconds.count() << std::endl;

  std::cout << std::endl;

}

 

The time literals are constant expressions therefore they can be evaluated at compile time. I have time durations for a German school hour (line 12), for a short break (line 14), for a long break (line 15), for his way to the school (line 17), and his homework (line 18). The result of the calculation schoolDaysInSeconds (line 20) is available at compile time.

schoolDay

Impressed? I'm not. His daily duty is only about 7 1/2 hour.

What's next?

The accuracy of the time tick is given by the used clock. In C++ we have the clocks std::chrono::system_clock, std::chrono::steady_clock, and std::chrono::high_resolution_clock. The three clocks will be the topic of my next post.

 

 

 

 

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Tags: time

Comments   

0 #1 PhSucharee 2016-09-20 09:15
I think there is a typo. std::ratio should be a minute.
Quote
0 #2 Rainer Grimm 2016-09-20 16:21
Quoting PhSucharee:
I think there is a typo. std::ratio should be a minute.
Thanks. You are right. I fix it.
Quote
0 #3 accounting 101 2017-02-18 16:36
Keep this going please, great job!
Quote

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