Daniel Walker wrote:
>
> On Mon, 23 May 2005, Oleg Nesterov wrote:
>
> > So the very first node will be skipped, iteration will be out of order,
> > and you will have the plist's *head* as a last element (which is not
> > struct rt_mutex_waiter, of course).
>
> True, but the first node is the list head which must be static,
> It's not an actual list member.
No. Look, plist_for_each() is just list_for_each_entry() now.
So, the the first iteration will play with head->sp_node.next,
not with head! And then you are doing ->dp_node.next again in
plist_entry().
I'd suggest you to write simple programm, and test. I bet I'm
right.
> > new_sp_head:
> > itr_pl2 = container_of(itr_pl->dp_node.prev, struct plist, dp_node);
> > list_add(&pl->sp_node, &itr_pl2->sp_node);
> >
> > Why? Just list_add_tail(&pl->sp_node, itr_pl->sp_node), you don't
> > need itr_pl2 at all.
>
> Wouldn't work . What if itr_pl has 15 elements at it's priority?
15 or 0, I think it would work. Look, it is *new* priority, and
pl->prio < itr_pl->prio, so pl should stay just before itr_pl
in ->sp_node list. That is why list_add_tail(&pl->sp_node, itr_pl->sp_node)
is enough.
Oleg.
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