On 30/01/17 17:26, Eric Dumazet wrote:
On Thu, 2016-10-06 at 20:50 -0400, David Miller wrote:
From: Anoob Soman <[email protected]>
Date: Wed, 5 Oct 2016 15:12:54 +0100

If a socket has FANOUT sockopt set, a new proto_hook is registered
as part of fanout_add(). When processing a NETDEV_UNREGISTER event in
af_packet, __fanout_unlink is called for all sockets, but prot_hook which was
registered as part of fanout_add is not removed. Call fanout_release, on a
NETDEV_UNREGISTER, which removes prot_hook and removes fanout from the
fanout_list.

This fixes BUG_ON(!list_empty(&dev->ptype_specific)) in netdev_run_todo()

Signed-off-by: Anoob Soman <[email protected]>
Applied and queued up for -stable, thanks.
This commit (6664498280cf "packet: call fanout_release, while
UNREGISTERING a netdev")
looks buggy :

We end up calling fanout_release() while holding a spinlock
( spin_lock(&po->bind_lock); )

But fanout_release() grabs a mutex ( mutex_lock(&fanout_mutex) ), and
this is absolutely not valid while holding a spinlock.

Yes, that is wrong.


Anoob, can you cook a fix, I guess you have a way to reproduce the thing
that wanted a kernel patch ?

(Please build your test kernel with CONFIG_LOCKDEP=y)

Sure, I am planning to move fanout_release(sk) after spin_unlock(bind_lock). Something like this.
                                }
                                if (msg == NETDEV_UNREGISTER) {
                                        packet_cached_dev_reset(po);
-                                       fanout_release(sk);
                                        po->ifindex = -1;
                                        if (po->prot_hook.dev)
dev_put(po->prot_hook.dev);
                                        po->prot_hook.dev = NULL;
                                }
                                spin_unlock(&po->bind_lock);
+                               if (msg == NETDEV_UNREGISTER) {
+                                       fanout_release(sk);
+                               }
                        }
                        break;

I will quickly test it out.

Thanks.


Thanks,
Anoob.

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