X-Authentication-Warning: acp3bf.physik.rwth-aachen.de: broeker owned process doing -bs Date: Mon, 20 Mar 2000 17:44:08 +0100 (MET) From: Hans-Bernhard Broeker X-Sender: broeker AT acp3bf To: Eli Zaretskii cc: djgpp-workers AT delorie DOT com Subject: Re: Unnormals??? In-Reply-To: Message-ID: MIME-Version: 1.0 Content-Type: TEXT/PLAIN; charset=US-ASCII Reply-To: djgpp-workers AT delorie DOT com Errors-To: dj-admin AT delorie DOT com X-Mailing-List: djgpp-workers AT delorie DOT com X-Unsubscribes-To: listserv AT delorie DOT com Precedence: bulk On Mon, 20 Mar 2000, Eli Zaretskii wrote: > > On Mon, 20 Mar 2000, Hans-Bernhard Broeker wrote: > > The sign bit alone does not identify the 'real indefinite'. The mantissa > > is also fixed. So the 'flag', if any, would be the whole 64 bits of > > information, not just the sign bit. > > Not really: if you flip the sign but leave the rest alone, the real > indefinite becomes a QNaN; you don't need to change the rest of the > bits. Well, the same happens if you manipulate any of the mantissa bits. So they are exactly as necessary for the thing to be the 'real indefinite' as the sign bit is. Eli, it seems as if just because the real indefinite has its sign bit set, by choice of Intel, you're trying to convince us that the sign bit of a NaN had better be neglected completely. Given that the only 'special' thing about real indefinite is that it is the only NaN that the FPU ever generates out of the blue, that looks like you're putting to much weight on this one item. Real indefinite, once present in a variable, or found in an FPU register, is treated exactly like all other QNaNs. I.e. it only has a different origin, but no different behaviour. Hans-Bernhard Broeker (broeker AT physik DOT rwth-aachen DOT de) Even if all the snow were burnt, ashes would remain.