Date: Wed, 24 Sep 1997 11:16:15 -0600 From: "Justin T. Gibbs" <gibbs@plutotech.com> To: Nate Williams <nate@mt.sri.com> Cc: "Justin T. Gibbs" <gibbs@plutotech.com>, current@freebsd.org Subject: Re: new timeout routines Message-ID: <199709241716.LAA24576@pluto.plutotech.com> In-Reply-To: Your message of "Wed, 24 Sep 1997 11:13:27 MDT." <199709241713.LAA12839@rocky.mt.sri.com>
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>> 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. It took the first entry off the list. The NetBSD timeout.9 page lists this as a bug. >Nate -- Justin T. Gibbs =========================================== FreeBSD: Turning PCs into workstations ===========================================
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