quartz: Fix deadlock with transform filter

[Alexandre Julliard]

"Maarten Lankhorst" <m.b.lankhorst at gmail.com> writes:

> It won't violate the locking rules, since this is the way it is
> designed. If you look at OutputPin_SendSample, you will see it takes
> the critical section to obtain whether the pin is connected to
> something, and if it is, it will send the data further. It does a
> great job of doing it safely, by obtaining all data under lock and
> adding a reference so that it can't die while receiving. The data is
> received by the receiving pin which takes a lock while processing the
> data, so the code is thread-safe.

I'm not saying it's not thread-safe, I'm saying it's wrong to release a
lock obtained by a caller. When you read the caller's code and you see
an Enter/Leave pair you expect that the lock is held all along, you
can't have inferior functions release it. The code needs to be
restructured so that Enter/Leave are properly matched.

-- 
Alexandre Julliard
julliard at winehq.org