Re: lockdep trace from posix timers

[Oleg Nesterov]

On 08/20, Peter Zijlstra wrote:
>
> On Mon, 2012-08-20 at 17:41 +0200, Oleg Nesterov wrote:
>
> > IMHO, this is just more natural.
>
> Depends on what you're used to I guess ;-)

I have to agree ;)

> > For example. keyctl_session_to_parent() does _cancel only to protect
> > from exploits doing keyctl(KEYCTL_SESSION_TO_PARENT) in an endless
> > loop. It could simply do task_work_add(), but in this case we need
> > fifo for correctness.
>
> I'm not entirely sure I see, not doing the cancel would delay the free
> until the executing of key_change_session_keyring()? doing that keyctl()
> in an indefinite loop involves going back to userspace, so where's the
> resource issue?

But the child does task_work_add(current->parent). IOW, there are 2
different tasks. Now suppose that ->parent sleeps.

> Also, I'm not seeing where the FIFO requirement comes from.

Again, suppose that ->parent sleeps. The last KEYCTL_SESSION_TO_PARENT
request should "win". Although I agree, this is not that important.

> > > Iterating a single linked queue in fifo
> > > seems more expensive than useful.
> >
> > Currently the list is fifo (we add to the last element), this is O(1).
>
> depends on what way you look at the list I guess, with a single linked
> list there's only one end you can add to in O(1), so we're calling that
> the tail?

Sorry, can't understand...

task->task_works points to the last node in the circular single-linked list,
task_work_add() adds the new element after the last one and updates
task->task_works. This is O(1).

> > But the list should be short, we can reverse it in _run() if we change
> > task_work_add() to add to the head.
>
> Reversing a (single linked) list is O(n^2)..

Hmm. This is O(n). You can simply iterate over this list once, changing
the ->next pointer to point back.

> which is indeed doable for
> short lists, but why assume its short?

I agree, it is better to not do this.

Oleg.

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