On Tue, Feb 18, 2014 at 07:58:49PM -0500, Waiman Long wrote:
> On 02/18/2014 04:37 PM, Peter Zijlstra wrote:
> >On Tue, Feb 18, 2014 at 02:39:31PM -0500, Waiman Long wrote:
> >>>>+ /*
> >>>>+ * At the head of the wait queue now
> >>>>+ */
> >>>>+ while (true) {
> >>>>+ u32 qcode;
> >>>>+ int retval;
> >>>>+
> >>>>+ retval = queue_get_lock_qcode(lock,&qcode, my_qcode);
> >>>>+ if (retval> 0)
> >>>>+ ; /* Lock not available yet */
> >>>>+ else if (retval< 0)
> >>>>+ /* Lock taken, can release the node& return */
> >>>>+ goto release_node;
> >>>>+ else if (qcode != my_qcode) {
> >>>>+ /*
> >>>>+ * Just get the lock with other spinners waiting
> >>>>+ * in the queue.
> >>>>+ */
> >>>>+ if (queue_spin_trylock_unfair(lock))
> >>>>+ goto notify_next;
> >>>Why is this an option at all?
> >>>
> >>>
> >>Are you referring to the case (qcode != my_qcode)? This condition will be
> >>true if more than one tasks have queued up.
> >But in no case should we revert to unfair spinning or stealing. We
> >should always respect the queueing order.
> >
> >If the lock tail no longer points to us, then there's further waiters
> >and we should wait for ->next and unlock it -- after we've taken the
> >lock.
> >
>
> A task will be in this loop when it is already the head of a queue and is
> entitled to take the lock. The condition (qcode != my_qcode) is to decide
> whether it should just take the lock or take the lock & clear the code
> simultaneously. I am a bit cautious to use queue_spin_trylock_unfair() as
> there is a possibility that a CPU may run out of the queue node and need to
> do unfair busy spinning.
No; there is no such possibility. Add BUG_ON(idx>=4) and make sure of
it.
There's simply no more than 4 contexts what can nest at any one time:
task context
softirq context
hardirq context
nmi context
And someone contending a spinlock from NMI context should be shot
anyway.
Getting more nested spinlocks is an absolute hard fail.
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