2025W Operating Systems (CS-3520-01) Course Materials & Test Answers | moodle31.upei.ca

Recall the code below from class where we used a condition variable to make sure the child runs before the parent prints. Why did we include the int variable done if signalling and waiting is done automatically with condition variables?

int done = 0;
pthread_mutex_t m = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t c = PTHREAD_COND_INITIALIZER;
void thr_exit() {
  Pthread_mutex_lock(&m);
  done = 1;
  Pthread_cond_signal(&c);
  Pthread_mutex_unlock(&m);
}
void *child(void *arg) {
  printf("child");
  thr_exit();
  return NULL;
}
void thr_join() {
  Pthread_mutex_lock(&m);
  while (done == 0)
    Pthread_cond_wait(&c, &m);
  Pthread_mutex_unlock(&m);
}
int main(int argc, char *argv[]) {
  printf("parent: begin");
  pthread_t p;
  Pthread_create(&p, NULL, child, NULL);
  thr_join();
  printf("parent: end");
  return 0;
}

If the child runs immediately on another CPU core so that it runs concurrently with the thread then there is a race condition for acquiring the lock m that having the parent first check done will prevent.

❌

If the child runs immediately after is created and we did not use done, then the child will signal before the parent puts itself to sleep. When the parent does sleep, it will then sleep forever since it missed its signal.

✅

This was a buggy implementation of the code that introduced a race condition for setting and checking done that is unnecessary since Pthread_cond_wait and Pthread_cond_signal perform the equivalent of checking and setting done, but atomically.

❌

If the child does not run immediately after is created (i.e., the parent continues running) and we did not use done, then the parent will go to sleep still holding the lock m and therefore prevent the child from being able to signal it. The parent will therefore sleep forever and the child will wait forever.

❌

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2025W Operating Systems (CS-3520-01)

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