netgraph(4) divert(4) to UDP Tunnel
From: Crist J. Clark (cristjc_at_comcast.net)
Date: 11/13/03
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Date: Thu, 13 Nov 2003 12:24:35 -0800 To: net@freebsd.org
I'm trying to play around with netgraph(4) for the first time and
there seem to be some aspects of it that haven't "clicked" in my head
just yet.
What I want to do seems like it should be pretty easy. I want to
send some packets through a UDP tunnel. There is an
/usr/share/examples/netgraph/udp.tunnel file that is close to what I
want, but not quite. I want to send packets that have been divert(4)ed
to the tunnel.
I can make my two ng_ksocket(8) nodes via the ngctl(8) interface,
+ mkpeer ksocket d0 inet/dgram/udp
+ name d0 udptun
+ msg d0 bind inet/192.168.64.70:10000
+ msg d0 connect inet/192.168.64.50:10000
+ mkpeer ksocket d1 inet/raw/divert
+ name d1 divtun
+ msg d1 bind inet/0.0.0.0:8668
But how do I then connect the two of them up? I assume that I use
'connect' within ngctl(8), but I haven't figured out what the
arguments need to be with the documentation and examples I've found.
The other thing I suspect I should be doing, is actually running the
'mkpeer' through the first node I create in ngctl(8), but I can't seem
to get that to work,
+ mkpeer ksocket d0 inet/dgram/udp
+ name d0 udptun
+ msg d0 bind inet/192.168.64.70:10000
+ msg d0 connect inet/192.168.64.50:10000
+ mkpeer d0 ksocket d1 inet/raw/divert
ngctl: send msg: Socket is already connected
I think it is actually complaining about the hook between my ngctl
node and the udptun node and not the creation of the divert socket?
Basically, I think my conceptual problem is with the fact that you
start with the ngctl(8) node in the middle of everything. How do I
create my new nodes and get the ngctl(8) node out of the middle?
--
Crist J. Clark | cjclark@alum.mit.edu
| cjclark@jhu.edu
http://people.freebsd.org/~cjc/ | cjc@freebsd.org
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