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Date:      Thu, 17 Feb 2000 11:44:13 +0100 (CET)
From:      Zahemszky Gabor <zgabor@CoDe.hu>
To:        freebsd-security@freebsd.org
Subject:   Re: ipfw - ipf
Message-ID:  <200002171045.LAA00428@CoDe.hu>
In-Reply-To: <E12Kbin-000PN1-00@f1.mail.ru> from Andrew Kopeyko at "Feb 15, 0 09:44:29 am"

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Hi!

> > a) if I have both ipf and ipfw in my kernel, which is the flow of a packet?
> > in -> ipf -> ipfw -> kernel | kernel -> ipfw -> ipf -> out
> > or the other?  (I used to use ipfw, and I'd like to switch - or learn - ipf.)
>  
> It depends on the order you modload their lkm's - earlier loaded will be "closer" to kernel.

Hm.  Can I load ipf dinamically?  Btw on 3.x, lkms are switched to klds, and
I didn't find a way to generate kld from ipf.

Every time, I compiled them into the kernel.

> But why do you use such strange thing? I had to used this for a week when i was switching from ipfw to ipf without interrupting clients services

For example, I like DUMMYNET and the bandwidth limiting with it.  But if I know
well, it's only available from ipfw's pipe commands.  Or is it possible
from ipf, or are there any other mechanism like dummynet?

> > b) Are there any ipfw to ipf converter?  I'd like to use (or write) it.
> 
> There is an addition to ipf -  'flc', 'filter language compiler'. It can compile some ruleset written in his own simple language to various firewalls rulesets: ipf, ipfw, CISCO, fwadmin, etc. See http://coombs.anu.edu.au/ipfilter/ for details.

Yes, I know it.  But I'd like to _convert_ my actual rules, and not to rewrite
them in another language.  (Of course, I can rewrite them in ipf's own
language :-)

By,

ZGabor at CoDe dot HU

-- 
#!/bin/ksh
Z='21N16I25C25E30, 40M30E33E25T15U!' ;IFS=' ABCDEFGHIJKLMNOPQRSTUVWXYZ ';set $Z ;for i { [[ $i = ? ]]&&print $i&&break;[[ $i = ??? ]]&&j=$i&&i=${i%?};typeset -i40 i=8#$i;print -n ${i#???};[[ "$j" = ??? ]]&&print -n "${j#??} "&&j=;typeset +i i;};IFS=' 0123456789 ';set $Z;X=;for i { [[ $i = , ]]&&i=2;[[ $i = ?? ]]||typeset -l i;X="$X $i";typeset +l i;};print "$X"


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