这是溢出吗？

问题描述

我有这段代码：

struct timeval start, end;
gettimeofday(&start, NULL);
//code I'm timing
gettimeofday(&end, NULL);
long elapsed = ((end.tv_sec-start.tv_sec)*1000000 + end.tv_usec-start.tv_usec);
ofstream timeFile;
timeFile.open ("timingSheet.txt");
timeFile << fixed << showpoint;
timeFile << setprecision(2);
timeFile << "Duration: " << elapsed << "\n";
timeFile.close();

哪个会输出已经通过的微秒数。但是，如果我更改这一行

Which will output the number of microseconds that has passed. However, if I change this line

long elapsed = ((end.tv_sec-start.tv_sec)*1000000 + end.tv_usec-start.tv_usec);

到：

long elapsed = ((end.tv_sec-start.tv_sec)*1000000 + end.tv_usec-start.tv_usec)/1000000.0;

我得到一个负值。为什么会发生这种情况？

I get a negative value. Why does this happen?

推荐答案

我使用从SO的某个地方借来的计时课。

I use a timing class that I borrowed from somewhere here on SO.

#include <time.h>
#include <sys/time.h>
#include <iomanip>
#include <iostream>

using namespace std;

class Timer
{
private:

timeval startTime;

public:

  void start()
  {
    gettimeofday(&startTime, NULL);
  }

  double stop()
  {
    timeval endTime;
    long seconds, useconds;
    double duration;

    gettimeofday(&endTime, NULL);

    seconds  = endTime.tv_sec  - startTime.tv_sec;
    useconds = endTime.tv_usec - startTime.tv_usec;

    duration = seconds + useconds/1000000.0;

    return duration;
  }

  static void printTime(double duration)
  {
    cout << setprecision(6) << fixed << duration << " seconds" << endl;
  }
};

例如：

Timer timer = Timer();
timer.start();
long x=0;
for (int i = 0; i < 256; i++)
  for (int j = 0; j < 256; j++)
    for (int k = 0; k < 256; k++)
      for (int l = 0; l < 256; l++)
        x++;
timer.printTime(timer.stop());

产生 11.346621秒。

对于我的，我得到：

Number of collisions: 0
Set size: 16777216
VM: 841.797MB
22.5810500000 seconds

这篇关于这是溢出吗？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！