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社区首页 >问答首页 >如何确保等待来自两个不同定时器的事件的线程不会错过任何事件？

问如何确保等待来自两个不同定时器的事件的线程不会错过任何事件？
EN

Stack Overflow用户
提问于 2014-01-08 09:05:16
回答 2查看 283关注 0票数 1

我想要创建一个线程，它等待由两个定时器在过期时发出信号的事件。当线程接收到信号时，它必须进行一些图像处理。我从下面的代码开始。(这是线程显示消息而不是进行图像处理的非常基本的实现)

代码语言：javascript
运行
复制
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/signal.h>
#include <sys/time.h>


struct itimerspec ts1,ts2;
struct sigevent se1,se2;

timer_t timerId1,timerId2;

pthread_cond_t imageProcessCond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t imageProcessMutex = PTHREAD_MUTEX_INITIALIZER;

void* imageProcessThread()
{
  int status = 0;
  int i = 0;

  while(i<2){

    status = pthread_mutex_lock (&imageProcessMutex);
    if(status != 0){
      perror("pthread_mutex_lock\n");
      exit(status);
    }

    status = pthread_cond_wait (&imageProcessCond, &imageProcessMutex);
    if (status != 0){
      perror("pthread_cond_wait\n");
      exit(status);
    }

    printf("signal received %s\n",__TIME__);

    i++;

    status = pthread_mutex_unlock (&imageProcessMutex);
    if(status != 0){
    perror("pthread_mutex_unlock\n");
    exit(status);
    }

  }

}

void timerFunc1 ()
{
  int status = 0;

  status = pthread_mutex_lock (&imageProcessMutex);
  if (status != 0){
    perror("pthread_mutex_lock\n");
    exit(EXIT_FAILURE);
  }

  printf("timer1 took lock %s\n",__TIME__);

  status = pthread_cond_signal (&imageProcessCond);
  if (status != 0){
    perror("pthread_cond_signal\n");
    exit(EXIT_FAILURE);
  }

  printf("timer1 signalled %s\n",__TIME__);

  status = pthread_mutex_unlock (&imageProcessMutex);
  if (status != 0){
    perror("pthread_mutex_unlock\n");
    exit(status);
  }

  printf("timer1 released lock %s\n",__TIME__);
}

void timerFunc2 ()
{
  int status = 0;

  status = pthread_mutex_lock (&imageProcessMutex);
  if (status != 0){
    perror("pthread_mutex_lock\n");
    exit(EXIT_FAILURE);
  }

  printf("timer2 took lock %s\n",__TIME__);

  status = pthread_cond_signal (&imageProcessCond);
  if (status != 0){
    perror("pthread_cond_signal\n");
    exit(EXIT_FAILURE);
  }

  printf("timer2 signalled %s\n",__TIME__);

  status = pthread_mutex_unlock (&imageProcessMutex);
  if (status != 0){
    perror("pthread_mutex_unlock\n");
    exit(status);
  }

  printf("timer2 released lock %s\n",__TIME__);
}

int main()
{

  int status = 0;
  pthread_t tImageProcessId;

  se1.sigev_notify = SIGEV_THREAD;
  se1.sigev_value.sival_ptr = &timerId1;
  se1.sigev_notify_function = timerFunc1;
  se1.sigev_notify_attributes = NULL;

  ts1.it_value.tv_sec = 20;
  ts1.it_value.tv_nsec = 0;
  ts1.it_interval.tv_sec = 20;
  ts1.it_interval.tv_nsec = 0;

  se2.sigev_notify = SIGEV_THREAD;
  se2.sigev_value.sival_ptr = &timerId2;
  se2.sigev_notify_function = timerFunc2;
  se2.sigev_notify_attributes = NULL;

  ts2.it_value.tv_sec = 20;
  ts2.it_value.tv_nsec = 0;
  ts2.it_interval.tv_sec = 20;
  ts2.it_interval.tv_nsec = 0;

  status = pthread_create((&tImageProcessId), NULL, &imageProcessThread, NULL);
  if(status!= 0){
    perror("pthread_create\n");
    exit(EXIT_FAILURE);
  }

  status = timer_create(CLOCK_REALTIME, &se1, &timerId1);
  if (status != 0){
    perror("timer_create\n");
    exit(EXIT_FAILURE);
  }

  status = timer_create(CLOCK_REALTIME, &se2, &timerId2);
  if (status != 0){
    perror("timer_create\n");
    exit(EXIT_FAILURE);
  }

  status = timer_settime(timerId1, 0, &ts1, 0);
  if (status != 0){
    perror("timer_create\n");
    exit(EXIT_FAILURE);
  }

  status = timer_settime(timerId2, 0, &ts2, 0);
  if (status != 0){
    perror("timer_create\n");
    exit(EXIT_FAILURE);
  }

sleep(80);

}

我已经为相同的时间间隔配置了两个定时器(我只是为了测试目的模拟了这个条件)，在实际场景中，定时器将被配置为不同的时间间隔，但在这种情况下，它们可能同时到期)。

我的理解是：

  1. 线程接受锁，进入等待状态，内部释放锁。
  2. timer1和timer2几乎同时到期，并且都试图取下锁。让我们假设timer1成功地获得了锁，计时器2保持在等待状态。
  3. 拥有锁的Timer1向事件发送信号并释放锁
  4. 等待事件的线程接受锁，显示消息并释放锁，然后再次进入等待状态。
  5. Timer2被计时器1保持在等待状态，接收锁，向事件发送信号并释放锁。

但是当我运行上面的代码时，线程实际上缺少了由timer1发出信号的事件。

这是我得到的输出

timer1锁11:40:20

timer1信号: 11:40:20

timer2锁11:40:20

timer2信号: 11:40:20

收到信号11:40:20

timer2释放锁11:40:20

timer1释放锁11:40:20

由timer1发出信号的事件正在消失。一旦timer1释放锁，计时器2就会获得锁，并且线程不会被执行。

如何确保控件从timer1到线程再到timer2？还有其他方法来实现这一点吗？请帮我解决这个问题

提前感谢

你好，阿皮塔

c
pthreads
EN

回答 2

Stack Overflow用户

发布于 2014-01-08 13:15:49

imageProcessThread函数存在同步中断。从imageProcessMutex互斥锁到下一个锁(第二个“同时”迭代)，代码不同步。可能第二个通知是从TimerFunc2函数发送的，当imageProcessThread不等待imageProcessCond时，这个通知就不会发生了。另一件重要的事情是虚假的唤醒。即使pthread_cond_wait函数没有发送通知，TimerFunc也可能会醒来。在您的情况下(根据输出)，它不会发生，但可能发生在其他条件下，应该得到适当的处理:当通知(真实的或虚假的)到达pthread_cond_wait时，再次检查谓词。

您的代码可以稍加修改以修复上面的漏洞。其主要思想是在函数之间共享局部变量i，让计时器增量它，并完全保护“main”等待循环。

代码语言：javascript
运行
复制
struct itimerspec ts1,ts2;
struct sigevent se1,se2;

timer_t timerId1,timerId2;

pthread_cond_t imageProcessCond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t imageProcessMutex = PTHREAD_MUTEX_INITIALIZER;
int i = 0;

void* imageProcessThread()
{
  int status = 0;
  //int i = 0;

  status = pthread_mutex_lock (&imageProcessMutex);
  while(i<2){
    //status = pthread_mutex_lock (&imageProcessMutex);
    status = pthread_cond_wait (&imageProcessCond, &imageProcessMutex);
    printf("signal received %s\n",__TIME__);
    //i++;
    //status = pthread_mutex_unlock (&imageProcessMutex);
  }
  status = pthread_mutex_unlock (&imageProcessMutex);
}

void timerFunc1 ()
{
  int status = 0;

  status = pthread_mutex_lock (&imageProcessMutex);
  i++;
  printf("timer1 took lock %s\n",__TIME__);

  status = pthread_cond_signal (&imageProcessCond);
  printf("timer1 signalled %s\n",__TIME__);

  status = pthread_mutex_unlock (&imageProcessMutex);
  printf("timer1 released lock %s\n",__TIME__);
}

void timerFunc2 ()
{
  int status = 0;

  status = pthread_mutex_lock (&imageProcessMutex);
  i++;
  printf("timer2 took lock %s\n",__TIME__);

  status = pthread_cond_signal (&imageProcessCond);
  printf("timer2 signalled %s\n",__TIME__);

  status = pthread_mutex_unlock (&imageProcessMutex);
  printf("timer2 released lock %s\n",__TIME__);
}

int main()
{

  int status = 0;
  pthread_t tImageProcessId;

  se1.sigev_notify = SIGEV_THREAD;
  se1.sigev_value.sival_ptr = &timerId1;
  se1.sigev_notify_function = timerFunc1;
  se1.sigev_notify_attributes = NULL;

  ts1.it_value.tv_sec = 20;
  ts1.it_value.tv_nsec = 0;
  ts1.it_interval.tv_sec = 20;
  ts1.it_interval.tv_nsec = 0;

  se2.sigev_notify = SIGEV_THREAD;
  se2.sigev_value.sival_ptr = &timerId2;
  se2.sigev_notify_function = timerFunc2;
  se2.sigev_notify_attributes = NULL;

  ts2.it_value.tv_sec = 20;
  ts2.it_value.tv_nsec = 0;
  ts2.it_interval.tv_sec = 20;
  ts2.it_interval.tv_nsec = 0;

  status = pthread_create((&tImageProcessId), NULL, &imageProcessThread, NULL);
  status = timer_create(CLOCK_REALTIME, &se1, &timerId1);
  status = timer_create(CLOCK_REALTIME, &se2, &timerId2);
  status = timer_settime(timerId1, 0, &ts1, 0);
  status = timer_settime(timerId2, 0, &ts2, 0);


sleep(80);

}

另一个次要通知是关于

代码语言：javascript
运行
复制
__TIME__

这是一个编译时间，因此您的输出将在每一行中显示相同的时间值。

我还建议阅读有关障碍同步的内容。

票数 1
EN

Stack Overflow用户

发布于 2014-01-09 09:18:15

谢谢你的答复

即使有上面的代码，我也错过了一个事件。我需要一个接一个地接收这些事件。

如果在锁定imageProcessMutex之前再使用一个锁，那么它工作正常。这两个事件都由图像处理线程接收。请看下面的代码。

代码语言：javascript
运行
复制
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/signal.h>
#include <sys/time.h>

struct itimerspec ts1,ts2;
struct sigevent se1,se2;

timer_t timerId1,timerId2;

pthread_cond_t imageProcessCond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t imageProcessMutex = PTHREAD_MUTEX_INITIALIZER;

pthread_mutex_t mainMutex = PTHREAD_MUTEX_INITIALIZER;

void* imageProcessThread()
{
  int status = 0;
  int i =0;

  status = pthread_mutex_lock (&imageProcessMutex);

  while(i<2){

    status = pthread_cond_wait (&imageProcessCond, &imageProcessMutex);

    printf("signal received %s\n",__TIME__);

    i++;
  }

  status = pthread_mutex_unlock (&imageProcessMutex);
}

void timerFunc1 ()
{
  int status = 0;

  status = pthread_mutex_lock (&mainMutex);

  status = pthread_mutex_lock (&imageProcessMutex);

  printf("timer1 took image process lock %s\n",__TIME__);

  status = pthread_cond_signal (&imageProcessCond);

  printf("timer1 signalled %s\n",__TIME__);

  status = pthread_mutex_unlock (&imageProcessMutex);

  printf("timer1 released image process lock %s\n",__TIME__);

  status = pthread_mutex_unlock (&mainMutex);

}

void timerFunc2 ()
{
  int status = 0;

  status = pthread_mutex_lock (&mainMutex);

  status = pthread_mutex_lock (&imageProcessMutex);

  printf("timer2 took image process lock %s\n",__TIME__);

  status = pthread_cond_signal (&imageProcessCond);

  printf("timer2 signalled %s\n",__TIME__);

  status = pthread_mutex_unlock (&imageProcessMutex);

  printf("timer2 released image lock %s\n",__TIME__);

  status = pthread_mutex_unlock (&mainMutex);

}

int main()
{

  int status = 0;
  pthread_t tImageProcessId;

  se1.sigev_notify = SIGEV_THREAD;
  se1.sigev_value.sival_ptr = &timerId1;
  se1.sigev_notify_function = timerFunc1;
  se1.sigev_notify_attributes = NULL;


  ts1.it_value.tv_sec = 20;
  ts1.it_value.tv_nsec = 0;
  ts1.it_interval.tv_sec = 0;
  ts1.it_interval.tv_nsec = 0;

  se2.sigev_notify = SIGEV_THREAD;
  se2.sigev_value.sival_ptr = &timerId2;
  se2.sigev_notify_function = timerFunc2;
  se2.sigev_notify_attributes = NULL;

  ts2.it_value.tv_sec = 20;
  ts2.it_value.tv_nsec = 0;
  ts2.it_interval.tv_sec = 0;
  ts2.it_interval.tv_nsec = 0;

  status = pthread_create((&tImageProcessId), NULL, &imageProcessThread, NULL);

  status = timer_create(CLOCK_REALTIME, &se1, &timerId1);

  status = timer_create(CLOCK_REALTIME, &se2, &timerId2);

  status = timer_settime(timerId1, 0, &ts1, 0);

  status = timer_settime(timerId2, 0, &ts2, 0);

  sleep(80);
}

我得到了以下输出

timer1拍摄了图像处理锁14:14:28

timer1信号14:14:28

收到信号14:14:28

timer1发布图像处理锁14:14:28

timer2拍摄了图像处理锁14:14:28

timer2信号14:14:28

timer2发布图像锁14:14:28

收到信号14:14:28

有了这个，我就不会错过任何一个事件。

我只想知道这是否可以在不使用pthread_cond_wait和pthread_cond_signal的情况下实现。

此外，您所提到的Barrier同步主要用于在多个线程(2个或多个线程)之间实现同步。但是在我的例子中，只有一个线程在等待event.So，我不确定障碍同步是否是正确的方法。请参考链接。

http://geekswithblogs.net/jolson/archive/2009/02/09/an-intro-to-barrier.aspx

感谢并致以问候

阿皮塔

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：

https://stackoverflow.com/questions/20991097

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