Date: Sun, 19 Mar 2000 11:14:53 +0200 (IST)
From: Eli Zaretskii <eliz AT is DOT elta DOT co DOT il>
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To: Eric Rudd <rudd AT cyberoptics DOT com>
cc: djgpp-workers AT delorie DOT com
Subject: Re: Unnormals???
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On Fri, 17 Mar 2000, Eric Rudd wrote:

> There is the cryptic statement in the Pentium Pro manual, vol. 2, 7.6:
> 
>      "The sign bit of a NaN is not interpreted."
> 
> What do you suppose they mean by that?

I interpret this as meaning that the sign bit of a NaN is a flag, not a 
sign.

> I had already done the experiment to produce the table in my earlier message;
> I only incorrectly labeled +NaN as SNaN.  Here is a hacky little program to
> try these things:
[snip]
>             if (xys[i][1] == '$') {    /* Motorola hex format */
>                xys[i][1] = xys[i][0];
>                sscanf(&xys[i][1], "%llx", &nd);
>             } else if (xys[i][0] == '$') {
>                sscanf(&xys[i][1], "%llx", &nd);
>             } else {                   /* C hex format, decimal, or octal */
>                sscanf(xys[i], "%lli", &nd);
>             }
>             x[i] = *((double *) &nd);

Beware of this last line: latest versions of GCC could do various tricks 
with it, which makes loading literal infinities and nans not trivial.  
You had better compiled without -O and examined the machine code.

> On my Pentium II, I found that -NaN = -NaN * -NaN; all other sign
> combinations produced +NaN.

This contradicts the Intel manual.  (So what else is new?)

>        220When  applied  to  infinite and NaN values, the -, +, and
>           space flag characters have their usual meaning; the # and
>           0 flag characters have no effect.
> 
> it seems that "real indefinite" must be printed out as negative.

Hmm?  How do you see it from this footnote?  IIRC it refers to the - and 
+ flags in the format specifier, which we do support.

The issue at hand is do we need to print the negative sign when the 
format is simply %f, say.