[pjsip] How to hangup a call - the right way ??

Turnaev Eugeny turnaev at t72.ru
Wed May 21 23:23:29 EDT 2008

On Wednesday 21 May 2008 18:32:04 Benny Prijono wrote:
> Well that's the problem in general with SIP, that is we need
> cooperation from remote party to achieve things. For example, if we
> send CANCEL or BYE, and remote party responds with "Sorry you can't do
> that", then what? Should we delete the call or not? In pjsip we chose
> the later since it tries to reflect the current state of the session,
> rather than what the user intends.
> But that's just a general statement, it's possible that this is a bug
> in pjsip instead (make sure you have the latest SVN version in this
> case). If you have a logging (at level 5) for your scenario then we
> may be able to find a specific solution.

For now i only have my app log and pjsua log level was 1.
Well, i will set up log level 5 in a next month when i will start my app again
i am sure bug calls will happen again, and then i will send pjsua log and mine
to this mail list. 

> >
> > 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?
> Well that *exactly* is the problem, isn't it? Imagine two threads are
> executing the same piece of code above, one thread comes from pjsip
> callback and is holding pjsip mutex and trying to acquire self.lock
> mutex, and the other thread is holding self.lock mutex and trying to
> acquire pjsip mutex, you end up with deadlock. This is what the past
> discussions were all about.

I see. So self.lock only causes deadlock and protects nothing
because of underlying mutexes in C.
Should i better remove self.lock.acquire() on those chunks of code?
I cant get - how should i deal with 2 threads calling pjsua functions,
without knowing if there is an mutex acquired in C level?
How to avoid deadlock from python?
Is there a way to find out if pjsua already acquired lock on mutex (from python level)?

More information about the pjsip mailing list