Question

我正在尝试编写一个能够使用C ++中的QueryPerformanceCounter计时事件的类。我们的想法是你创建一个计时器对象，给一个函数一个双倍格式的时间，并计算直到那个时间过去，然后再做东西。理想情况下，此类将用于计算游戏中的时间（例如，计时器在一秒钟内计数60次）。当我编译这段代码时，它只是打印0到控制台，似乎永远。但我注意到了一些我无法理解的错误。如果我单击控制台窗口的滚动条并按住它，则计时器实际上正确计数。如果我输入5.0，例如，然后快速点击并按住滚动条5秒或更长时间，当我放手时程序将打印“完成!!!＆＃39;”。那么，当我让它将经过的时间打印到控制台时，为什么它没有正常计数呢？是否有打印到控制台的故障，或者我的计时代码有问题？以下是代码：

   #include <iostream>
#include <iomanip>
#include "windows.h"

using namespace std;



int main()
{

    setprecision(10); // i tried to see if precision in the stream was the problem but i don't think it is

    cout << "hello! lets time something..." << endl;

    bool timing = 0;                // a switch to turn the timer on and off
    LARGE_INTEGER T1, T2;           // the timestamps to count
    LARGE_INTEGER freq;             // the frequency per seccond for measuring the difference between the stamp values

    QueryPerformanceFrequency(&freq); // gets the frequency from the computer

    // mil.QuadPart = freq.QuadPart / 1000; // not used


    double ellapsedtime = 0, desiredtime; // enter a value to count up to in secconds
    // if you entered 4.5 for example, then it should wait for 4.5 secconds

    cout << "enter the amount of time you would like to wait for in seconds (in double format.)!!" << endl;
    cin >> desiredtime;

    QueryPerformanceCounter(&T1);   // gets the first stamp value
    timing = 1;                    // switches the timer on

    while(timing)
    {
        QueryPerformanceCounter(&T2);      // gets another stamp value
        ellapsedtime += (T2.QuadPart - T1.QuadPart) / freq.QuadPart;   // measures the difference between the two stamp
        //values and then divides them by the frequency to get how many secconds has ellapsed
        cout << ellapsedtime << endl;
        T1.QuadPart = T2.QuadPart; // assigns the value of the second stamp to the first one, so that we can measure the
        // difference between them again and again

        if(ellapsedtime>=desiredtime) // checks if the elapsed time is bigger than or equal to the desired time,
        // and if it is prints done and turns the timer off
        {
            cout << "done!!!" << endl;
            timing = 0; // breaks the loop
        }
    }



    return 0;
}

Answer 1

您应该在ellapsedtime中存储自fisrt调用QueryPerformanceCounter以来经过的微秒数，并且您不应该覆盖第一个时间戳。

工作代码：

// gets another stamp value
QueryPerformanceCounter(&T2);     

// measures the difference between the two stamp
ellapsedtime += (T2.QuadPart - T1.QuadPart);

cout << "number of tick " << ellapsedtime << endl;
ellapsedtime *= 1000000.;
ellapsedtime /= freq.QuadPart;
cout << "number of Microseconds " << ellapsedtime << endl;

// checks if the elapsed time is bigger than or equal to the desired time
if(ellapsedtime/1000000.>=desiredtime)         {
   cout << "done!!!" << endl;
   timing = 0; // breaks the loop
}

请务必阅读：Acquiring high-resolution time stamps

C ++计时事件控制台bug？

1 个答案: