On Thursday 23 March 2006 10:57, Florian Effenberger wrote:
> Hi,

> >> eth0  IP: 85.123.123.10

> > Let's say "IP A"

> >> eth0:1  IP: 85.123.123.15

> > "IP B"

> > That could be due to a network hang fixed _after_ 2.4.25 - upgrade to my
> > 2.4.x -bs2 patchset on my homepage.

> I run 2.6.8. :-)

IIRC that has that bug - anyway I would never recommend a 2.6.8 UML kernel.

> > host: tap0 uses "IP A", eth0:1 is removed.
> >
> > UML: eth0 is assigned "IP B".

> Stop - that won't work. IP A is reserved for my original host, the UML
> device should be reachable *only* with IP B. IP A = original host, IP B
> = virtual host/UML.

It's what I'm doing, so use what I suggested.

You're misunderstanding a fact: if tap0 receives a packet with its IP, it's 
for the host; so tap0 should have IP A. To send a packet to IP B (UML) IP B 
must be the destination address (set inside UML), not the source address (set 
on tap0).

If UML sends a packet to IP A, it will go to host tap0; if anybody else sends 
a packet to IP A, it will arrive to host eth0.

The fact that UML has IP B is set via proxy ARP. If you use uml_net, it's done 
automatically, otherwise you do it by hand via proxy ARP, see the site 
"Virtual Networking" page.
-- 
Inform me of my mistakes, so I can keep imitating Homer Simpson's "Doh!".
Paolo Giarrusso, aka Blaisorblade (Skype ID "PaoloGiarrusso", ICQ 215621894)
http://www.user-mode-linux.org/~blaisorblade

                
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