On Thu, 24 Jul 2008, Nick Piggin wrote:
quoted text
> Hey, something kind of cool (and OT) I've just thought of that we can
> do with ticket locks is to take tickets for 2 (or 64K) nested locks,
> and then wait for them both (all), so the cost is N*lock + longest spin,
> rather than N*lock + N*avg spin.

Isn't this deadlocky?

E.g. one task takes ticket x=1, then other task comes in and takes x=2
and y=1, then first task takes y=2.  Then neither can actually
complete both locks.

Miklos
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