Mail Archives: djgpp-workers/2000/03/20/12:03:41
On Mon, 20 Mar 2000, Eli Zaretskii wrote:
>
> On Mon, 20 Mar 2000, Hans-Bernhard Broeker wrote:
>
> > Eli's point of view, as I read it: No, it is not negative, because the
> > term 'negative' does not have a valid mathematical meaning for NaNs.
>
> 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.
> > My point against this reasoning by Eli is that the word 'negative', if
> > found in the C99 standard, does not necessarily mean the same as the word
> > 'negative' in mathematics.
>
> IIRC, the standard doesn't say what does it mean by ``negative''.
No, it doesn't, at least not anywhere in the text. I haven't checked
the standards it cites for definition of terms, though. E.g. it cites
'ISO/IEC 2382-1:1993' on
Information technology -- Vocabulary --
Part 1: Fundamental terms.
That document may contain a definition of 'negative', for the context of
computers and floating point hardware. Or the IEC 60559 one on floating
point arithmetics (already referenced before, in this discussion) may.
If 'negative' really is not defined by the standard, this may imply that,
from the C99 standard's point of view, we would be allowed to treat sign
bits in NaNs as nonexistant. I'd hate that decision, but the standard
seems to allow it.
Hans-Bernhard Broeker (broeker AT physik DOT rwth-aachen DOT de)
Even if all the snow were burnt, ashes would remain.
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