使用sleep()时计算执行时间

问题描述

我正在尝试从运行Angstrom Linux发行版的Beagle Bone中的ADC读取数据.我需要使用类似sleep()的延迟机制，以便仅在特定时间读取样本，以帮助符合特定的采样率. 我还需要计算执行时间.

I am trying to read data from the ADC in the Beagle Bone, running Angstrom Linux distribution. I need to use a delay mechanism like sleep(), to only read samples at specific time, to help conform to a specific sample rate. I also am required to calculate the execution time.

以下是示例POC(概念验证)，以演示我面临的问题:

Here is a sample POC (proof of concept), to demonstrate the problem I am facing:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


int main()
{
    clock_t begin, end;

    while(1)
    {
        begin = clock();
        sleep(1); // delay for 1 sec
        end = clock();
        printf("Execution time = %f\n",((float)(end-begin)/CLOCKS_PER_SEC));
    }
}

我总是以0.000000作为执行时间.

I always get the execution time as 0.000000.

为什么我没有得到1.000000秒的结果?我的猜测是调用sleep()将抢占我的程序，但我不确定.

Why do I not get my result as 1.000000 seconds? My guess is calling sleep() will pre-empt my program, but I am not sure.

我还需要什么其他选项来计算经过的执行时间(包括延迟)?

What other option do I have to calculate elapsed execution time which includes a delay?

推荐答案

计算执行时间的解决方案是在程序的开头和结尾获取时间戳.然后有所作为.

The solution to calculate execution time is to get the timestamp at the beginning of your program and at the end. Then make the difference.

#include <stdio.h>
#include <time.h>

int main() {

    time_t begin;
    time(&begin);

  // Somethings

   time_t end;
   time(&end);

  printf("Execution time %f\n", difftime(end, begin));
  return (0);
}

#include <stdio.h>
#include <time.h>
#include <sys/time.h>

int main() {

  struct timeval  tv;
  gettimeofday(&tv, NULL);

double begin =
  (tv.tv_sec) * 1000 + (tv.tv_usec) / 1000 ;


 sleep(2);

 gettimeofday(&tv, NULL);

double end =
  (tv.tv_sec) * 1000 + (tv.tv_usec) / 1000 ;

  printf("Execution time %f\n", end - begin);
  return (0);
}

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