On Tue, 28 Mar 2006, Thomas Gleixner wrote:
> On Tue, 2006-03-28 at 22:17 +0100, Esben Nielsen wrote:
> > I think we talk about the situation
>
> No, we talk about existing lock chains L(0) --> L(n).
>
> > B locks 1 C locks 2 D locks 3
> > B locks 2, boosts C and block
> > A locks 2
> > A is boost B
> > A drop it's spinlocks and is preempted
> > C unlocks 2 and auto unboosts
> > B is running
> > B locks 3, boosts C and blocks
> > A gets a CPU again
> > A boosts B
> > A boosts D
> >
> > Is there anything wrong with that?
> > And in the case where A==D there indeed is a deadlock which will be
> > detected.
>
> If you get to L(x) the underlying dependencies might have changed
> already as well as the dependencies x ... n. We might get false
> positives in the deadlock detection that way, as a deadlock is an
> "atomic" state.
As I see it you might detect a circular lock graph "atomically". But is
that a "deadlock"? Yes, if you rule out signals and timeouts, this
situation does indeed deadlock your program.
But if you count in signals and timeouts your algoritm also gives "false
positives": You can detect a circular lock but when you return from
rt_mutex_slowlock(), a signal is delivered and there is no longer a
circular dependency and most important: The program wouldn't be
deadlocked even if you didn't ask for deadlock detection and your task in
that case would block.
I would like to see an examble of a false deadlock. I don't rule them out
in the present code. But they might be simple to fix.
Esben
>
> tglx
>
>
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