on 14/05/2008 18:17 Andriy Gapon said the following:
> I am trying the small attached program on FreeBSD 6.3 (amd64,
> SCHED_4BSD) and 7-STABLE (i386, SCHED_ULE), both with libthr as threads
> library and on both it produces "BROKEN" message.
>
> I compile this program as follows:
> cc sched_test.c -o sched_test -pthread
>
> I believe that the behavior I observe is broken because: if thread #1
> releases a mutex and then tries to re-acquire it while thread #2 was
> already blocked waiting on that mutex, then thread #1 should be "queued"
> after thread #2 in mutex waiter's list.
>
> Is there any option (thread scheduler, etc) that I could try to achieve
> "good" behavior?
>
> P.S. I understand that all this is subject to (thread) scheduler policy,
> but I think that what I expect is more reasonable, at least it is more
> reasonable for my application.
>
>


Daniel Eischen has just kindly notified me that the code (as an
attachment) didn't make it to the list, so here it is inline.


#include
#include
#include
#include


pthread_mutex_t mutex;
int count = 0;

static void * thrfunc(void * arg)
{
while (1) {
pthread_mutex_lock(&mutex);

count++;
if (count > 10) {
fprintf(stderr, "you have a BROKEN thread scheduler!!!\n");
exit(1);
}

sleep(1);

pthread_mutex_unlock(&mutex);
}
}

int main(void)
{
pthread_t thr;

#if 0
pthread_mutexattr_t attr;
pthread_mutexattr_init(&attr);
pthread_mutexattr_settype(&attr, PTHREAD_MUTEX_RECURSIVE_NP);
pthread_mutex_init(&mutex, &attr);
#else
pthread_mutex_init(&mutex, NULL);
#endif

pthread_create(&thr, NULL, thrfunc, NULL);

sleep(2);

pthread_mutex_lock(&mutex);
count = 0;
printf("you have good thread scheduler\n");
pthread_mutex_unlock(&mutex);

return 0;
}

--
Andriy Gapon
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