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Date: Mon, 2 Jul 2001 08:25:38 -0700
From: "Crist J. Clark" <cristjc@earthlink.net>
To: David Malone <dwmalone@maths.tcd.ie>
Cc: David Hill <djhill@novagate.net>, freebsd-current@FreeBSD.ORG
Subject: Re: syslogd and -a
Message-ID: <20010702082538.B448@blossom.cjclark.org>
Reply-To: cjclark@alum.mit.edu
References: <20010701234125.7a7d3e3a.djhill@novagate.net> <20010701212044.Q296@blossom.cjclark.org> <20010702093842.A13480@walton.maths.tcd.ie>
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In-Reply-To: <20010702093842.A13480@walton.maths.tcd.ie>; from dwmalone@maths.tcd.ie on Mon, Jul 02, 2001 at 09:38:42AM +0100
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On Mon, Jul 02, 2001 at 09:38:42AM +0100, David Malone wrote:
> On Sun, Jul 01, 2001 at 09:20:44PM -0700, Crist J. Clark wrote:
> > Hmmm... Looks like,
> > 
> >   # syslogd -a 192.168.1.0/29
> > 
> > Will work and,
> > 
> >   # syslogd -a 192.168.1.1/29
> > 
> > Won't.
> 
> That's the standard behaviour of a netmask, isn't it? The usual
> way to check if host h is in network/netmask n/m is to check if:
> 
> 	(h & m == n)
> 
> this means that the bits of the network which are not in the mask
> must be zero.

That's exactly what happens in the syslogd(8) code. However, I think
that should be,

  n &= m
  .
  .
  .
  ((h & m) == n)

That is, why allow the user to enter a network number that is not
/really/ the network number? Either flag an error or do the
calculation for the user. I think doing the calculation is the more
sensible choice. Commiting it to CURRENT now.
-- 
Crist J. Clark                           cjclark@alum.mit.edu

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