Why does pthread_cond_signal not work?

Question

I am currently learing all around POSIX threads (pthread).

I now have created a simple program which increased a shared value by 7 until above 10000 then it should signal a condition to the next thread which decreases it by 3 until under 1000. At last it should divide the result through 2 and main should output the result.

My code:

pthread_t threads[3];
pthread_cond_t cond_a, cond_b;
pthread_mutex_t mutex;

int counter;

void * worker_one();
void * worker_two();
void * worker_three();

int main(int argv, const char ** argc) {
    counter = 0;

    pthread_cond_init(&cond_a, NULL);
    pthread_cond_init(&cond_b, NULL);
    pthread_mutex_init(&mutex, NULL);

    pthread_create(&threads[0], NULL, worker_one, NULL);
    pthread_create(&threads[1], NULL, worker_two, NULL);
    pthread_create(&threads[2], NULL, worker_three, NULL);

    pthread_join(threads[0], NULL);
    pthread_join(threads[1], NULL);
    pthread_join(threads[2], NULL);

    printf("Value started at %d and ends with %d.\n", 0, counter);

    return 0;
}

void * worker_one() {
    printf("Worker one started.\n");

    pthread_mutex_lock(&mutex);

    printf("Worker one starting work.\n");
    while (counter < 10000) {
        counter += 7;
    }

    pthread_cond_signal(&cond_a);

    printf("Worker one finished work with: %d.\n", counter);

    pthread_mutex_unlock(&mutex);
    pthread_exit(NULL);
}

void * worker_two() {
    printf("Worker two started.\n");

    pthread_mutex_lock(&mutex);
    pthread_cond_wait(&cond_a, &mutex);

    printf("Worker two starting work.\n");
    while (counter > 1000)
        counter -= 3;

    printf("Worker two finished work with: %d.\n", counter);

    pthread_cond_signal(&cond_b);
    pthread_mutex_unlock(&mutex);

    sleep(1);

    pthread_exit(NULL);
}

void * worker_three() {
    printf("Worker three started.\n");

    pthread_mutex_lock(&mutex);
    pthread_cond_wait(&cond_b, &mutex);

    printf("Worker three starting work.\n");

    counter /= 2;

    printf("Worker three finished work with: %d.\n", counter);

    pthread_mutex_unlock(&mutex);
    pthread_exit(NULL);
}

For some reason the whole execution hangs around the first thread. The signal is also fired but thread two does not react.

Can somebody tell me what I am doing wrong?

I've moved the code from gist to SO. Don't just link to code on an external site, please include relevant code in the question. — user7116, May 29 '13 at 16:25

score 5 · Accepted Answer · edited May 23 '17 at 12:22

5

I have answered a similar question here: pthread condition variables on Linux, odd behaviour.

The problem is that you wait before even testing the condition you want to wait for is true. What happens is that thread 1 signals before thread 2 is waiting, therefore the signal is lost and thread 2 will be waiting forever.

In order to avoid this, first test what you want to wait for, then wait only if it's not here.

EDIT: Ok, here is a possible solution with only one mutex and one condtion (untested)

Thread 1:

pthread_mutex_lock(&mutex); 
while(thread_1_should_work == false){ // wait until the condition is satisfied
  pthread_cond_wait(&cond, &mutex); 
}

//at this point, we owe the mutex and we know thread_1_should_work is true; 

// do work 

thread_1_shoudl_work = false; 
thread_2_should_work = true; 

pthread_cond_broadcast(&cond); //wake up any waiting thread (if it's not their turn, they'll call wait again)
pthread_mutex_unlock(&mutex);

... and so on.

edited May 23 '17 at 12:22

Community

1
1

answered May 29 '13 at 16:31

Ben

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Is the testing case a simple external flag or something pthread internal? I use the example from https://computing.llnl.gov/tutorials/pthreads but there are no flags actually – bodokaiser May 29 '13 at 16:33
@bodokaiser: you'll have to make up a variable, but a simple boolean flag will do. Spurious wakeups are a reality unfortunately. – user7116 May 29 '13 at 16:35
@sixlettervariables so the main idea of conditions are just some more specific mutex signaling (with out the actual locking). Is this correct? – bodokaiser May 29 '13 at 16:36
Yes, in your case is thread_is_working flag would work (except that it has more than 6 letters :) I think the answer that the linked answer can help. – Ben May 29 '13 at 16:38
Actually, wait() releases the mutex and yield the thread atomically so that nothing happens while the thread is being yield. – Ben May 29 '13 at 16:41
@Ben Ok, I understand. Are semaphores actually the secure tool to make the signaling between threads (instead of using external flags)? – bodokaiser May 30 '13 at 08:00
@bodokaiser Conditions are perfectly secure, I would stick to them and use flags for each thread to indicate whether it should be working or not. – Ben May 30 '13 at 08:17

Why does pthread_cond_signal not work?

1 Answers1