How to provide a sequence of interleaving threads to show that a code breaks and doesn't provide perfect synchronization?

Question

I know what the following code does and I know why it is a broken code for synchronization as it has only one conditional variable while we need two but I don't know how to provide a sequence of interleaving threads for showing it doesn't work. Can you show why this code doesn't work with an example?

1   cond_t cond = PTHREAD_COND_INITIALIZER; 
2  mutex_t mutex=PTHREAD_MUTEX_INITIALIZER;;
3
4  void *producer(void *arg) {
5    int i;
6    for (i = 0; i < loops; i++) {
7      Pthread_mutex_lock(&mutex);
8      while (count == 1)
9        Pthread_cond_wait(&cond, &mutex);
10     put(i);
11     Pthread_cond_signal(&cond);
12     Pthread_mutex_unlock(&mutex);
13   }
14 }
15
16 void *consumer(void *arg) {
17   int i;
18   for (i = 0; i < loops; i++) {
19     Pthread_mutex_lock(&mutex);
20     while (count == 0)
21       Pthread_cond_wait(&cond, &mutex);
22     int tmp = get();
23     Pthread_cond_signal(&cond);
24     Pthread_mutex_unlock(&mutex);
25     printf("%d\n", tmp);
26   }
27 }

Answer 1

Presuming that put() sets count to 1 and get() sets count to 0, this code is actually fine as long as you have only one producer and one consumer.

If you have more than one producer, then the pthread_cond_signal() in the producer might wake up one of the other producers instead of a consumer, and then no consumers will proceed. The same problem exists if you have more than one consumer.

You can fix this either by introducing separate condition variables for empty and full, or by using pthread_cond_broadcast() instead of pthread_cond_signal().

Answer 2

"cond" is not initialized. You have a race condition between consumer and producer. If any of the "while" condition becomes true, they will result in infinite loop as no instruction changes the values of "count" in the "while" scopes (both).