In-Reply-To: <s5hejxb7xrb.wl%[email protected]>
On Tue, 27 Jun 2006 11:28:56 +0200, Takashi Iwai wrote:
> > > > The code in question is doing..
> > > >
> > > > __list_add(&deleted_list,
> > > > client->ports_list_head.prev,
> > > > client->ports_list_head.next);
> > > >
> > > > which looks fishy, as those two elements aren't going to be consecutive,
> > > > as __list_add expects.
> > >
> > > I think the code behaves correctly but probably misusing __list_add().
> > > It movies the whole entries from an existing list_head A
> > > (clients->ports_list_head) to a new list_head B (deleted_list).
> > > The above is exapnded:
> > >
> > > A->next->prev = B;
> > > B->next = A->next;
> > > B->prev = A->prev;
> > > A->prev->next = B;
> > >
> > > Any better way to achieve it using standard macros?
> >
> > Why can't you just list_move() the elements ?
>
> No, list_move() can't move the whole elements without loop.
>
> A solution is
>
> list_add(B, A);
> list_del_init(A);
>
> (although this introduces a bit more code :)
Shouldn't it be like this?
ports_list_first = client->ports_list_head.next;
list_del_init(client->ports_list_head);
list_splice(ports_list_first, &deleted_list);
--
Chuck
"You can't read a newspaper if you can't read." --George W. Bush
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