On Thu, 2012-03-08 at 19:28 +0100, Peter Zijlstra wrote:
> On Thu, 2012-03-08 at 13:23 -0500, Steven Rostedt wrote:
> > So basically what you tried to do was just set the owner of the lock to
> > have the priority of the task that wants the lock, until it releases it?
> > But by doing it without having this task sleep?
>
> No, by having it sleep ;-)
>
That was the second part of my email.
> So you do the full PI sleeping lock thing, except you return fail if you
> loose the acquisition race on wakeup and you mark this waiter as
> 'special'.
>
> Then on every rt_mutex block you have to do a deadlock analysis on the
> PI blocking chain (preferably shared with PI boost traversal of said
> chain), during that scan you collect all special tagged waiters.
>
> If you find a deadlock, wake all these special waiters and have them
> return -EDEADLK.
>
> I guess you could also do the full spin_deadlock() and do away with the
> try part and purely rely on the deadlock detection.
But do you release the lock first? For example, we have:
@@ -410,7 +411,7 @@ static inline struct dentry *dentry_kill
if (inode && !spin_trylock(&inode->i_lock)) {
relock:
seq_spin_unlock(&dentry->d_lock);
- cpu_relax();
+ cpu_chill();
return dentry; /* try again with same dentry */
}
By doing the test at the trylock, we can easily hit the deadlock,
because we still hold dentry->d_lock. But by moving the block to the
cpu_chill(), then we are less likely to hit the deadlock.
Perhaps call it, cpu_chill_on_lock().
-- Steve
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