On 22/06/15(Mon) 17:12, Alexander Bluhm wrote:
> On Wed, Jun 17, 2015 at 02:08:14PM +0200, Martin Pieuchot wrote:
> > diff -u -p -r1.340 if.c
> > --- net/if.c 16 Jun 2015 11:09:39 -0000 1.340
> > +++ net/if.c 17 Jun 2015 12:03:36 -0000
> > @@ -530,6 +530,15 @@ if_input_process(void *xmq)
> > continue;
> > }
> >
> > +#if NBRIDGE > 0
> > + if (ifp->if_bridgeport && (m->m_flags & M_PROTO1) == 0) {
> > + m = bridge_input(m);
> > + if (m == NULL)
> > + continue;
> > + }
> > + m->m_flags &= ~M_PROTO1; /* Loop prevention */
> > +#endif
>
> Should we reset the loop prevention only if our call to bridge_input()
> did set M_PROTO1? Something like this
>
> if (ifp->if_bridgeport && (m->m_flags & M_PROTO1) == 0) {
> m = bridge_input(m);
> if (m == NULL)
> continue;
> m->m_flags &= ~M_PROTO1; /* Loop prevention */
> }
Yes and no :)
bridge_input() will set M_PROTO1 on the mbuf copies that it enqueues on
its ports.
If you receive a packet on em0 in bridge0 with tun0, you want to call
bridge_input() only once, but you'll call if_input() in em0 and tun0.
So the first packet will enter if_input() without M_PROTO1, go through
bridge_input() then be processed by the stack. Then the copy of this
packet created in bridge_input() will have the M_PROTO1 flag set and
when it will be dequeued by if_input() it won't be passed to
bridge_input() again.
Does that make sense?
