Question

我想在代码片段（单独的函数）中收集程序的运行时，当前的策略是计算每个部分的执行时间（chrono::duration）并将它们加在一起。但我必须处理两种不同的情况（使用不同的输入调用两次函数），并使用静态变量timer来保持分隔的持续时间。现在我想在第二种情况之前重置变量。我怎样才能做到这一点？当然我可以使用两个连续的system_clock :: now（）的持续时间，但这似乎是不必要的。我尝试timer = 0或timer(0)，但无效。例如：

class myClass {
  static std::chrono::milliseconds timer_A;
  void foo();
  void bar();
}

在cpp文件中：

std::chrono::milliseconds timer_A(0);
foo() {
  ...
  // get someDuration1;
  timer_A += someDuration1;
  ....
}
bar() {
  ...
  // get someDuration2;
  timer_A += someDuration2;
  ...
  cout << timer_A.count();
}

有一个main()会两次拨打foo()和bar()。即，

int main() {
  if(someCondition) {
    foo();
    bar();
  }
  if(otherCondition) {
    // here I need reset timer_A to 0 s.t. record the runtime again, how        ??????????
    foo();
    bar();
  }

Answer 1

我的回答是 abit ，但这里有：

timer_A = std::chrono::miliseconds::zero();

您还可以创建持续时间模板，然后使用其Zero（）函数，这样您就不必知道测量的持续时间（见下文）。

Chrono没有自己的计时器类，但你可以很容易地创建一个：

//----------------------------------------------------------------
/* Timer
Simple timer class using C++11 chrono.
example use:
{
    Timer_t t; // use global typedef for standard timer.
    t.Start(); // start the timer.
    ... do something you want to time ...
    t.End();   // end the timer
    t.CalcDuration; // Calculate and store (inside the Timer class) the interval between start and end.
    cout << "Time in milliseconds: " << t.GetDuration().count() << end; // cout the interval
}
*/
template< typename precision = long double, typename ratio = std::milli >
class Timer final
{
public:
    //---------------- aliases ----------------
    using timeDuration_t = std::chrono::duration<precision, ratio>;
    using timePoint_t = std::chrono::time_point<std::chrono::system_clock, timeDuration_t>;
    using this_type = Timer< precision, ratio >;

    //---------------- special member functions ----------------
    // NOTE: all the default compiler generated member functions will suffice.

    //---------------- member functions ----------------

    //--------------------------------
    /* Start()
        Starts the timer.
    */
    inline void Start()
    {
        m_start = std::chrono::system_clock::now();
    }

    //--------------------------------
    /* End()
        Ends the timer.
    */
    inline void End()
    {
        m_end = std::chrono::system_clock::now();
    }

    //--------------------------------
    /* CalcDuration()
        Calculates the time elapsed (duration)
        in between the previous calls to Start and End().

        NOTE: make sure to have called Start() and End() before calling this function.
        NOTE: Start() and End() can be called in any order.
    */
    void CalcDuration()
    {
        m_duration = std::max( m_start, m_end ) - std::min( m_start, m_end );
    }

    //--------------------------------
    /* GetDuration()
        Returns the calculated duration.

        NOTE: make sure to call CalcDuration() before calling this function.
    */
    timeDuration_t const& GetDuration() const
    {
        return m_duration;
    }

    //--------------------------------
    /* Zero()
        Zeros the internal members, effectively resetting the Timer.
    */
    void Zero()
    {
        m_start = timeDuration_t::zero();
        m_end = m_start;
        m_duration = m_end;
    }

    //--------------------------------
    /* TestLatency( i_count )
        Tests the latency / error of the timer class.

        NOTE: this number is how inaccurate your timings can be.
    */
    static timeDuration_t TestLatency( size_t const i_count = 512 )
    {
        this_type t;
        timeDuration_t tSum = timeDuration_t::duration::zero();
        for( size_t i = 0; i < i_count; ++i )
        {
            t.Start();
            t.End();
            t.CalcDuration();
            tSum += t.GetDuration();
        }
        return tSum / i_count;
    }

private:
    //---------------- private member data ----------------
    timePoint_t     m_start;
    timePoint_t     m_end;
    timeDuration_t  m_duration;
};

using Timer_t = Timer<>;

Answer 2

如果是毫秒类型：timer_A = milliseconds(0);

如何重置chrono :: duration值？

2 个答案: