From owner-freebsd-current@FreeBSD.ORG Tue Oct 12 15:44:10 2004 Return-Path: Delivered-To: freebsd-current@freebsd.org Received: from mx1.FreeBSD.org (mx1.freebsd.org [216.136.204.125]) by hub.freebsd.org (Postfix) with ESMTP id A3B9716A4CE; Tue, 12 Oct 2004 15:44:10 +0000 (GMT) Received: from athena.softcardsystems.com (mail.softcardsystems.com [12.34.136.114]) by mx1.FreeBSD.org (Postfix) with ESMTP id 4993C43D41; Tue, 12 Oct 2004 15:44:10 +0000 (GMT) (envelope-from sah@softcardsystems.com) Received: from athena (athena [12.34.136.114])i9CGgWlq031724; Tue, 12 Oct 2004 11:42:32 -0500 Date: Tue, 12 Oct 2004 11:42:32 -0500 (EST) From: Sam X-X-Sender: sah@athena To: =?iso-8859-1?q?Dag-Erling_Sm=F8rgrav?= In-Reply-To: Message-ID: References: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed cc: freebsd-current@freebsd.org cc: freebsd-arch@freebsd.org Subject: Re: mbuf w/o pkthdr? X-BeenThere: freebsd-current@freebsd.org X-Mailman-Version: 2.1.1 Precedence: list List-Id: Discussions about the use of FreeBSD-current List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Tue, 12 Oct 2004 15:44:10 -0000 >> Are all packets supposed to have the M_PKTHDR flag? Why? > > IIRC, M_PKTHDR indicates the first mbuf in a chain when a packet is > split across multiple mbufs. This usually only happens for outgoing > packets, where protocol headers are constructed in separate mbufs > which are prepended to the chain as the packet moves down the stack. That's kind of my understanding (with some PACKET_TAG* stuff going on). I don't have split headers, though. Neither would an arp frame, but he too gets a packet header and fills it out. I don't mind following suit, I'm just wondering what the convention is for. Perhaps we use mbufs without packet headers for something special? Sam