Date: Wed, 24 Sep 1997 11:19:00 -0600 (MDT) From: Nate Williams <nate@mt.sri.com> To: "Justin T. Gibbs" <gibbs@plutotech.com> Cc: Nate Williams <nate@mt.sri.com>, current@freebsd.org Subject: Re: new timeout routines Message-ID: <199709241719.LAA12880@rocky.mt.sri.com> In-Reply-To: <199709241716.LAA24576@pluto.plutotech.com> References: <199709241713.LAA12839@rocky.mt.sri.com> <199709241716.LAA24576@pluto.plutotech.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. So, the old behavior (that we've all grown to love now that it's gone. :) was used for this many years, yet was full of bugs? Do hash collisions not ever occur 'in the current pre-CAM system'? Nate
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