X-Authentication-Warning: acp3bf.physik.rwth-aachen.de: broeker owned process doing -bs Date: Mon, 20 Mar 2000 18:20:42 +0100 (MET) From: Hans-Bernhard Broeker X-Sender: broeker AT acp3bf To: 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. If you flip any mantissa bit of it, the real indefinite becomes a QNaN; you don't have to change the sign bit. Where's the difference that would make the sign bit special, in that respect? The real indefinite is exactly one pattern of bits, per FP datatype. Its only reason for existence is that Intel felt they had to choose a single common result for all cases where a NaN was generated 'out of the blue', i.e. with no other NaN to copy the freely variable bits from that are allowed in an NaN. Once it has been generated, it's just an ordinary QNaN like all the others. Hans-Bernhard Broeker (broeker AT physik DOT rwth-aachen DOT de) Even if all the snow were burnt, ashes would remain.