> Bruce Momjian wrote:
> >
> > The unusual case here is that deadlock is not checked on request, but
> > only after waiting on the lock for a while. This is because deadlock
> > detection is an expensive operation.
>
> But that is what happens. If one of the locks is not obtained, the
> algorithm does wait on that lock (after releasing the other locks).
> In the case of a deadlock (tom's scenario #1) it would wait forever,
> but the deadlock detection will find it in there and break it.
OK, I thought you were talking about the LOCK a,b case, not the other
case where we had a previous LOCK statement. Sorry.
--
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