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Date:      Mon, 13 Aug 2001 16:56:04 +0300
From:      Peter Pentchev <roam@ringlet.net>
To:        default - Subscriptions <default013subscriptions@hotmail.com>
Cc:        freebsd-security@freebsd.org, freebsd-questions@freebsd.org
Subject:   Re: Easy IPFW question...
Message-ID:  <20010813165603.B1119@ringworld.oblivion.bg>
In-Reply-To: <OE26Wd7KKQpQq5pneeF0000b932@hotmail.com>; from default013subscriptions@hotmail.com on Mon, Aug 13, 2001 at 02:12:29AM -0500
References:  <OE26Wd7KKQpQq5pneeF0000b932@hotmail.com>

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On Mon, Aug 13, 2001 at 02:12:29AM -0500, default - Subscriptions wrote:
> Hi,
> 
> I'm kinda new to IPFW, and I was unable to figure this out by myself...
> 
> I want to block an I.P. range, say 192.168.0.1, with a netmask of
> 255.255.0.0 ...
> 
> The rule I tried was this:
> ipfw add deny log all from 192.168.0.1/16 to any via ed0

Try 192.168.0.0/16 - the bits that are zeroed in the netmask must be
also zeroed in the address.  This is so because of the way the address/netmask
calculations are performed: when an address, say 192.168.5.12, is tested
against a 192.168.0.0/16 combination, a bitwise 'and' operation is performed
between the address to check (192.168.5.12) and the netmask (255.255.0.0).
Then, the result - 192.168.0.0 - is compared to the network address that
you have specified.  Since the 'and' operation clears the last 16 bits,
a network address of 192.168.0.1 cannot match anything - it has its
last bit set.

Try 192.168.0.0/16, it will probably work.  If it fails, try
192.168.0.0/255.255.0.0.

G'luck,
Peter

-- 
I am the meaning of this sentence.

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