From owner-freebsd-ipfw Wed Mar 27 3:30:12 2002 Delivered-To: freebsd-ipfw@freebsd.org Received: from fm1.freemail.hu (fm1.freemail.hu [195.228.242.201]) by hub.freebsd.org (Postfix) with SMTP id EE0B237B416 for ; Wed, 27 Mar 2002 03:30:07 -0800 (PST) Received: (qmail 5050 invoked by uid 662851); 27 Mar 2002 12:30:05 +0100 Date: Wed, 27 Mar 2002 12:30:05 +0100 (CET) From: Szabados Jozsef To: freebsd-ipfw@freebsd.org 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 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 Hi! I would need some starting info about transparent proxying. 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? Like linux with iptables: getsockopt(... SO_ORIGINAL_DST ...) Or with ipfilter's natlookup_t structure. (Long to insert here) But how, with ipfw? Any help (man/rtfm ;-) is appreciated, Best regards, 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