Date: Sun, 2 Apr 2000 08:00:28 +0200 (IST)
From: Eli Zaretskii <eliz AT is DOT elta DOT co DOT il>
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To: Dieter Buerssner <buers AT gmx DOT de>
cc: djgpp-workers AT delorie DOT com
Subject: Re: Unnormals???
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On Sun, 26 Mar 2000, Dieter Buerssner wrote:

> I was not aware, 
> that even extended IEEE 754 format NaNs and Infinities must have the 
> most significant mantiassa bit set. (And I still wonder about the 
> intention of this requirement.)

I'm guessing that Intel didn't want the infinities and quiet NaNs to
produce the denormal exceptions, so they defined them all with the
integer bit set.

> >    > When converted to a long double, these two have the following bit
> >    > patterns:
> >    > 
> >    >  pos_nanshort = 7fff 0001 0000 0000
> >    >  neg_nanshort = ffff 0001 0000 0000
> > 
> >    Have your forgotten 0000 at the end of the previous two lines?
> > 
> > No, I haven't, but I don't see why do they matter.  A long double
> > number is only 10 bytes long.
> 
> Then, I don't understand, how I should read your bit patterns.
> There are only 16 hexadecimal digitits in your examples, while
> a long double should have 16 hex digits for the significand, and
> four hexdigits, when combining sign and exponent.

Sorry, I misunderstood you.  I thought you were talking about the last
(6th) element of the array in *your* code, not the missing 5th word in
*my* code.  I indeed erroneously omitted the 5th word.

> >    I think, they are not unnormals. I think this discussion has shown,
> >    that unnormals must have a finite exponent.
> > 
> > I don't see the finite exponent requirement in the Intel manual.
> 
> In the Intel manual, Table 7-12, (unsupported extended real 
> encodings) the biased exponent field for unnormals is given as:
> 
>  11..10
>    .
>  00..01
> 
> So, 00..00 and 11..11 are excluded.

Yes, the Intel manual calls them pseudo-NaNs.

> Yes. My example numbers should never be used.  But you have to print 
> something for them anyway, when they are produced by buggy code.

I suggest "unnormal", like it already does.