From owner-freebsd-current Wed Sep 24 10:13:33 1997 Return-Path: Received: (from root@localhost) by hub.freebsd.org (8.8.7/8.8.7) id KAA11380 for current-outgoing; Wed, 24 Sep 1997 10:13:33 -0700 (PDT) Received: from ns.mt.sri.com (SRI-56K-FR.mt.net [206.127.65.42]) by hub.freebsd.org (8.8.7/8.8.7) with ESMTP id KAA11375 for ; Wed, 24 Sep 1997 10:13:30 -0700 (PDT) Received: from rocky.mt.sri.com (rocky.mt.sri.com [206.127.76.100]) by ns.mt.sri.com (8.8.7/8.8.7) with ESMTP id LAA01753; Wed, 24 Sep 1997 11:13:28 -0600 (MDT) Received: (from nate@localhost) by rocky.mt.sri.com (8.7.5/8.7.3) id LAA12839; Wed, 24 Sep 1997 11:13:27 -0600 (MDT) Date: Wed, 24 Sep 1997 11:13:27 -0600 (MDT) Message-Id: <199709241713.LAA12839@rocky.mt.sri.com> From: Nate Williams MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7bit To: "Justin T. Gibbs" Cc: Nate Williams , current@freebsd.org Subject: Re: new timeout routines In-Reply-To: <199709241709.LAA24417@pluto.plutotech.com> References: <199709241656.KAA12715@rocky.mt.sri.com> <199709241709.LAA24417@pluto.plutotech.com> X-Mailer: VM 6.29 under 19.15 XEmacs Lucid Sender: owner-freebsd-current@freebsd.org X-Loop: FreeBSD.org Precedence: bulk > So you assume that regardless of what pointers the client gives you, > even if they give you the same pair twice without an intervening > expiration or untimeout call, that there will be no collisions in > the hash table? How did the original code in untimeout() determine what to pull off the table? Obviously there is enough information in the untimeout() call to uniquely determine which entry to use, and that same information was used in timeout(), so we must be able to build a perfect hash function. Nate