Subj : Re: pthread problems: wake up a thread via pthread_cond_signal()
To   : comp.programming.threads
From : David Schwartz
Date : Sun Jan 02 2005 03:21 pm


>    for(i=0; i<2; i++){
>        in_err = pthread_create(&in_thr[i], NULL, (void *(*)(void
> *))thr1,    (void *)i);
>     if (in_err)
>         exit (1);
>    }
>
>    out_err = pthread_create(&out_t2, NULL, (void *(*)(void *))thr2,
> NULL);
>    if (out_err)
>        exit (1);
>
>
>    return (0);
> }

    You fall off the end of 'main'. You need to join all your threads.

>       pthread_mutex_lock(&lock[id]);
>
>       if(status[id] == 1){
>        printf("wait %i\n", id);
>        pthread_cond_wait(&in_turn[id], &lock[id]);
>       }
>
>
>     pthread_mutex_lock(&status_lock);
>        work[id] = 1;
>     pthread_mutex_unlock(&status_lock);

    Notice that 'work[id]' can only be set to '1' by a thread that holds the 
'lock[id]' mutex.

>        pthread_mutex_lock(&lock[i]);
>
>        printf("---- write data from thread %i ---\n", i);
>
>    pthread_mutex_lock(&status_lock);
>        status[i] = 0;
>        pthread_mutex_unlock(&status_lock);
>
>  printf("(int)status[i] %i\n", status[i]);
>   printf("(int)work[i] %i\n", work[i]);
>
>
>    while((int)work[i] != 1){
>     pthread_cond_signal(&in_turn[i]);
>     printf("## signal to thread %i\n", i);
>    }

    So how is another thread going to set 'work[i]' to 1? It would have to 
hold the 'lock[i]' mutex, and *this* thread holds that mutex.

    DS

.