netgraph(4) divert(4) to UDP Tunnel

[Crist J. Clark]

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 at alum.mit.edu
                                   |     cjclark at jhu.edu
http://people.freebsd.org/~cjc/    |     cjc at freebsd.org