On Wednesday 21 May 2008 17:21:19 Benny Prijono wrote: > > > > I have a debug listing of my app where call to > > status = py_pjsua.call_hangup(call_id, 0, None, None) > > returned status => 0 but a call still alive and in active state. > > > > In my app right after py_pjsua.call_hangup(...), approx 100 lines of python code later > > app start a new call with call_make_call without waiting a callback to return that previous > > call is disconnected. From time to time it happens that call_hangup returns 0 but a call keeps > > living in active state .. state text EARLY.. and total duration increments. i can send my app log if it needed. > > > > There's no guarantee that call_hangup() will trigger DISCONNECTED > callback to be called immediately. In the worst case, you'll get the > callback up to 32 seconds after call_hangup(), if remote is not > responding. In this case, the call duration will be kept incremented > of course. > Thanks for answers, i will use DNDEBUG :) 32 seconds - well.. i already ended up 2 times with no calls available in pool (before set up cleaner of buggy calls), both times - total_duration of some calls was > 1200 seconds... ( but unfortunably i did not dumped detailed information about calls at that time, but i bet calls was active and in EARLY state ) So i would not say > There's no guarantee that call_hangup() will trigger DISCONNECTED > callback to be called immediately. i would say: There's no guarantee that call_hangup() will trigger DISCONNECTED at all. one should keep trying to hangup untill it dies :) > This has been discussed few times before, please see > http://www.google.com/custom?q=Timed-out+trying+to+acquire+PJSUA&domains=pjsip.org&sitesearch=pjsip.org > > Unfortunately we lost the list archive when we moved the mailing list > late last year, but you can still get it from Google cache. > I already googled with this and found nothing userfull. What i am saying: i am sure 100% that my code do NOT enters py_pjsua.call_get_info() 2 times.. i have isolated every call to py_pjsua.call_get_info() with semaphore. self.lock.acquire() try: call_info = py_pjsua.call_get_info(self.current_call) ... do some stuff finally: self.lock.release() So if i isolated every call to py_pjsua.call_get_info how can i came with pjsua_call.c Timed-out trying to acquire PJSUA mutex (possibly system has deadlocked) in pjsua_call_get_info() message in log? Maybe some other py_pjsua.* functions calling pjsua_call_get_info ?
