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 <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <pthread.h>
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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