On Wed, Oct 09, 2002 at 04:07:45PM -0700, Terry Lambert wrote:
> Stefan Farfeleder wrote:
> > Is it just a warning or does it pose a real problem?
> >
> > I think the problem with the current code is that knote_{en,de}queue can
> > be executed in parallel (on another CPU, spl*() can't prevent that, can
> > it?) with kqueue_scan and that kq->kq_head thus can be corrupted.
> > Or am I totally wrong?
>
> My patch would have worked, in that case, since it would ensure
> one marker entry with a unique stack address per simultaneous
> scanner.
>
> It has to be that the queue itself is being deleted out from
> under it: the problem is not the scan, nor the insert or the
> delete.
>
> Most likely, this is for an object whose queue is not tracked
> by process, or for a process queue that's being examined by
> another process (e.g. kevent's on fork/exit/etc.).
>
> You can verify this for your own satisfaction by looking at the
> pointer manipulation order for the insertion and deletion; the
> insertion sets the next pointer before setting the pointer to
> the inserted object, and the deletion sets the pointer that
> used to point to the deleted object to the delete object's next,
> before deleting the object. Thus, traversals in progress should
> not result in an error.
Imagine this scenario where CPU 0 inserts a knote kn1 (the marker) in
knote_scan and CPU 1 kn2 in kqueue_enqueue:
CPU 0 | CPU 1
--------------------------------+-------------------------------
kn1->kn_tqe.tqe_next = NULL; |
|
--------------------------------+-------------------------------
kn1->kn_tqe.tqe_prev = | kn2->kn_tqe.tqe_next = NULL;
kq_head.tqh_last; |
--------------------------------+-------------------------------
*kq_head.tqh_last = kn1; | kn2->kn_tqe.tqe_prev =
| kq_head.tqh_last;
--------------------------------+-------------------------------
kq_head.tqh_last = | *kq_head.tqh_last = kn2;
&kn1->kn_tqe.tqe_next; |
--------------------------------+-------------------------------
| kq_head.tqh_last =
| &kn2->kn_tqe.tqe_next;
The marker would never appear on the queue.
Regards,
Stefan Farfeleder
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