唤醒单个线程而不是在 pthreads 中忙等待

Question

我不确定标题是否反映了我在这里问的内容，但如果没有很长的标题，我能做到的最好。我正在尝试worker thread在pthreads. 我想从main函数中生成一组线程，然后main线程将作业委托给工作线程并等待所有线程完成，然后再为它们分配下一个作业（实际上，要求是将线程安排在一个块中，很像 CUDA编程模型，但在 CPU 上。虽然它与当前问题无关）。该job数组用于向每个线程指示作业类型。目前，我已经使用信号量实现了这一点，这会强制执行繁忙的等待。我正在寻找方法使线程仅在需要时才进入睡眠和唤醒状态，而不是连续轮询。

每个线程执行的函数

volatile int jobs[MAX_THREADS]; // global job indicator array
sem_t semaphore;                // semaphore to indicate completion
thread_execute(void *args)
{
  tid = get_id(args);
  while(jobs[tid] != -1)
  {
    if(jobs[tid] == 0) continue; // no job
    if(jobs[tid] == JOBS_1)
    {
      jobs1();
      jobs[tid] = 0; // go back to idle state
      sem_post(&semapahore);
    }
    if(jobs[tid] == JOBS_2)
    {
      jobs2();
      jobs[tid] = 0; // go back to idle state
      sem_post(&semapahore);
    }
  }

  pthread_exit(NULL);
}

主要功能如下

int main()
{
  sem_init(&semaphore, 0, 0);
  jobs[0...MAX_THREADS] = 0;
  spawn_threads();

  // Dispatch first job
  jobs[0...MAX_THREADS] = JOBS_1;
  int semvalue = 0;
  while (semvalue < MAX_THREADS) // Wait till all threads increment the semaphore
    sem_getvalue(&sempaphore, &semvalue);

  sem_init(&semaphore, 0, 0); // Init semaphore back to 0 for the next job
                              // I'm actually using diff. semaphores for diff. jobs
  jobs[0...MAX_THREADS] = JOBS_2;
  while (semvalue < MAX_THREADS)
    sem_getvalue(&sempaphore, &semvalue);

  jobs[0...MAX_THREADS] = -1; // No more jobs
  pthread_join();
}

此实现的问题在于main线程忙于等待所有工作线程完成，工作线程也在不断轮询作业数组以检查新作业。当线程进入睡眠状态并在需要时沿着信号处理程序和使用唤醒时，是否有更好的方法来执行此操作，pthread_kill()但使用单独的信号处理程序有点混乱。

Answer 1

您可以使用条件变量使线程进入休眠状态，直到收到信号。

volatile int jobs[MAX_THREADS]; // global job indicator array
pthread_cond_t th_cond;     // threads wait on this
pthread_mutex_t th_mutex;   // mutex to protect the signal
int busyThreads = MAX_THREADS;

pthread_cond_t m_cond;      // main thread waits on this
pthread_mutex_t m_mutex;    // mutex to protect main signal

thread_execute(void *args)
{
  tid = get_id(args);
  while(jobs[tid] != -1)
  {
    if(jobs[tid] == 0) continue; // no job
    if(jobs[tid] == JOBS_1)
    {
      jobs1();
      jobs[tid] = 0; // go back to idle state
      pthread_mutex_lock(&th_mutex);      
          pthread_mutex_lock(&m_mutex);   
          --busyThreads;                       // one less worker
          pthread_cond_signal(&m_cond);        // signal main to check progress
          pthread_mutex_unlock(&m_mutex);
      pthread_cond_wait(&th_cond, &th_mutex);   // wait for next job
      pthread_mutex_unlock(&th_mutex);      
    }
    if(jobs[tid] == JOBS_2)
    {
      jobs2();
      jobs[tid] = 0; // go back to idle state
      pthread_mutex_lock(&th_mutex);
      --busyThreads;
      pthread_cond_wait(&th_cond, &th_mutex);
      pthread_mutex_unlock(&th_mutex);
    }
  }

  pthread_exit(NULL);
}

然后在主要：

int main()
{
  sem_init(&semaphore, 0, 0);
  jobs[0...MAX_THREADS] = 0;
  spawn_threads();

  // Dispatch first job
  jobs[0...MAX_THREADS] = JOBS_1;
  int semvalue = 0;

  pthread_mutex_lock(&m_mutex);
  while(busyThreads > 0)        // check number of active workers
      pthread_cond_wait(&m_cond, &m_mutex);   
  pthread_mutex_unlock(&m_mutex);

  busyThreads = MAX_THREADS;
  pthread_mutex_lock(&th_mutex);
  pthread_cond_broadcast(&th_cond);   // signal all workers to resume
  pthread_mutex_unlock(&th_mutex);

  // same for JOBS_2;

  jobs[0...MAX_THREADS] = -1; // No more jobs
  pthread_join();
}