From owner-freebsd-ipfw Wed Mar 27 4:33: 4 2002 Delivered-To: freebsd-ipfw@freebsd.org Received: from fm2.freemail.hu (fm2.freemail.hu [195.228.242.202]) by hub.freebsd.org (Postfix) with SMTP id 5826337B419 for ; Wed, 27 Mar 2002 04:32:59 -0800 (PST) Received: (qmail 60129 invoked by uid 662851); 27 Mar 2002 13:32:46 +0100 Date: Wed, 27 Mar 2002 13:32:46 +0100 (CET) From: Szabados Jozsef Subject: transparent proxying To: freebsd-ipfw@freebsd.org In-Reply-To: Message-ID: X-Originating-IP: [212.24.188.125] X-HTTP-User-Agent: Mozilla/5.0 (X11; U; FreeBSD i386; en-US; rv:0.9.7) Gecko/20020123 MIME-Version: 1.0 Content-Type: TEXT/PLAIN; CHARSET=ISO-8859-2 Content-Transfer-Encoding: QUOTED-PRINTABLE Sender: owner-freebsd-ipfw@FreeBSD.ORG Precedence: bulk List-ID: List-Archive: (Web Archive) List-Help: (List Instructions) List-Subscribe: List-Unsubscribe: X-Loop: FreeBSD.ORG Sorry, I forgot the subject. Szabados Jozsef =EDrta: > Hi! >=20 > I would need some starting info about transparent proxying. >=20 > So, first I forward the packet, on port xx to 127.0.0.1:xxxx > the proxy get the packet, and it will see the destination ip > 127.0.0.1, isn't it? So how can I find out the original > destination > ip address of the packet? >=20 > Like linux with iptables: > getsockopt(... SO_ORIGINAL_DST ...) >=20 > Or with ipfilter's natlookup_t structure. > (Long to insert here) >=20 > But how, with ipfw? >=20 > Any help (man/rtfm ;-) is appreciated, > Best regards, >=20 > Szabados Jozsef > ps.: sorry for my bad english ;-) > To Unsubscribe: send mail to majordomo@FreeBSD.org with "unsubscribe freebsd-ipfw" in the body of the message