Hello Tom,

Thank you for the clarification!


Le lun. 9 avr. 2018 à 17:04, Tom Lane <t...@sss.pgh.pa.us> a écrit :

> Christophe Pettus <x...@thebuild.com> writes:
> >> On Apr 9, 2018, at 07:33, Thomas Poty <thomas.p...@gmail.com> wrote:
> >> ok, and long  answer ? is it random?
> > It's not literally random, but from the application point of view, it's
> not predictable.  For example, it's not always the one that opened first,
> or any other consistent measure.
> It's whichever one runs the deadlock detector first after the circular
> wait becomes established.  For instance:
> * Process A takes lock L1
> * Process B takes lock L2
> * Process A tries to take lock L2, blocks
> * Process B tries to take lock L1, blocks (now a deadlock exists)
> Process A will run the deadlock detector one deadlock_timeout after
> blocking.  If that happens before B has blocked, then A will see
> no deadlock and will go back to waiting.  In that case, when B's
> own deadlock_timeout expires and it runs the deadlock detector,
> it will see the deadlock and fix it by canceling its own wait.
> On the other hand, if B started to wait less than one deadlock_timeout
> after A did, then A will be first to observe the deadlock and it will
> cancel itself, not B.
> So you can't predict it unless you have a lot of knowledge about
> the timing of events.  You could probably make it more predictable
> by making deadlock_timeout either very short or very long, but
> neither of those are desirable things to do.
>                         regards, tom lane

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