On Wed, Jan 11, 2006 at 06:25:36PM +0100, Esben Nielsen wrote:
> So how many locks do we have to worry about? Two.
> One for locking the lock. One for locking various PI related data on the
> task structure, as the pi_waiters list, blocked_on, pending_owner - and
> also prio.
> Therefore only lock->wait_lock and sometask->pi_lock will be locked at the
> same time. And in that order. There is therefore no spinlock deadlocks.
> And the code is simpler.
Ok, got a question. How do deal with the false reporting and handling of
a lock circularity window involving the handoff of task A's BKL to another
task B ? Task A is blocked trying to get a mutex owned by task B, task A
is block B since it owns BKL which task B is contending on. It's not a
deadlock since it's a hand off situation.
I didn't see any handling of this case in the code and I was wondering
if the traversal logic you wrote avoids this case as an inherent property
and I missed that stuff ?
bill
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