Operating Systems: Virtualization, Concurrency & Persistence/

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Better, But Still Broken: While, Not If

Let's try to work on the solution of the producer/consumer problem from the last lesson.

We'll cover the following...

- Why this works?

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int loops; // must initialize somewhere...
cond_t  cond;
mutex_t mutex;
void *producer(void *arg) {
  int i;
  for (i = 0; i < loops; i++){
    Pthread_mutex_lock(&mutex);         // p1
    if (count == 1)                     // p2
      Pthread_cond_wait(&cond, &mutex); // p3
    put(i);                             // p4
    Pthread_cond_signal(&cond);         // p5
    Pthread_mutex_unlock(&mutex);       // p6
  }
} 
void *consumer(void *arg) {
 int i;
 for(i = 0; i < loops; i++) {
   Pthread_mutex_lock(&mutex);         // c1
   while (count == 0)                  // c2      
    Pthread_cond_wait(&cond, &mutex);  // c3
    int tmp = get();                   // c4
    Pthread_cond_signal(&cond);        // c5
    Pthread_mutex_unlock(&mutex);      // c6
    printf("%d\n", tmp);
  }
}

Why this works?

Think about why this works; now consumer $T_{c1}$ wakes up and (with the lock held) immediately re-checks the state of the shared variable (c2). If the buffer is empty at that point, the consumer simply goes back to sleep (c3). The corollary if is also changed to a while in the producer (p2).

Thanks to Mesa semantics, a simple rule to remember with condition variables is to always use while loops. Sometimes you don’t have to re-check the condition, but it is always safe to do so; just do it and be happy.

However, this code still has a bug, the second of two problems mentioned in the last lesson. Can you see it? It has something to do with the fact that there is only one condition variable. Try to figure out what the problem is, before reading ahead. DO IT!

(pause for you to think, or close your eyes…)