Date: Mon, 20 Mar 2000 18:31:59 +0200 (IST) From: Eli Zaretskii X-Sender: eliz AT is To: Hans-Bernhard Broeker 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, Hans-Bernhard Broeker wrote: > > Not only because of this, but also because how Intel treats the real > > indefinite. It is clear (to me) from that treatment that they use the > > sign bit as a flag, to the effect that this NaN was produced by an > > operation wher none of the operands was a NaN. > > 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.