Hi Lizardo.
* Anderson Lizardo <anderson.lizardo@xxxxxxxxxxxxx> [2011-04-04 15:22:29 -0400]:
> Hi Gustavo,
>
> On Fri, Apr 1, 2011 at 6:33 PM, Gustavo F. Padovan
> <padovan@xxxxxxxxxxxxxx> wrote:
> > @@ -203,13 +217,14 @@ static void __l2cap_chan_add(struct l2cap_conn *conn, struct sock *sk)
> > l2cap_pi(sk)->omtu = L2CAP_DEFAULT_MTU;
> > }
> >
> > - __l2cap_chan_link(l, sk);
> > + __l2cap_chan_link(l, chan);
> > }
> >
> > /* Delete channel.
> > * Must be called on the locked socket. */
> > -void l2cap_chan_del(struct sock *sk, int err)
> > +void l2cap_chan_del(struct l2cap_chan *chan, int err)
> > {
> > + struct sock *sk = chan->sk;
> > struct l2cap_conn *conn = l2cap_pi(sk)->conn;
> > struct sock *parent = bt_sk(sk)->parent;
> >
> > @@ -219,7 +234,7 @@ void l2cap_chan_del(struct sock *sk, int err)
> >
> > if (conn) {
> > /* Unlink from channel list */
> > - l2cap_chan_unlink(&conn->chan_list, sk);
> > + l2cap_chan_unlink(&conn->chan_list, chan);
> > l2cap_pi(sk)->conn = NULL;
> > hci_conn_put(conn->hcon);
> > }
> > @@ -253,6 +268,8 @@ void l2cap_chan_del(struct sock *sk, int err)
> > kfree(l);
> > }
> > }
> > +
> > + kfree(chan);
>
> Just wondering: isn't also necessary to do "l2cap_pi(sk)->chan =
> NULL;" here, supposing the underlying struct sock will still be
> allocated?
No, it is always allocated. Just check the the comment on the top of the
function "Must be called on the locked socket" or the code itself.
--
Gustavo F. Padovan
http://profusion.mobi
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