nan or -nan?

[Paul D. Anderson]

Walter Bright Wrote:

> On 11/17/2011 5:29 PM, bearophile wrote:
> > Walter:
> >
> >> On 11/17/2011 3:57 PM, bearophile wrote:
> >>> It's a NaN, but floating point designers have given a sign to NaNs for a
> >>> (small) purpose.
> >>
> >> What is that purpose?
> >
> > I didn't know the answer, so I've done a short research (finding texts like
> > this: http://grouper.ieee.org/groups/754/email/msg03893.html ), I have found
> > that languages usually ignore the sign of the NaNs (while the sign of zero
> > has some purposes). Producing the same bit patterns for 32 and 64 bit
> > compilers is useful for output binary consistency for comparisons, but
> > comparing floating point values bitwise is usually a not so wise thing to do.
> > So I think there is no need to put this in Bugzilla.
> 
> There is a NaN "payload" in the significand which it is polite to preserve, 
> though no program has ever been observed to make use of it. The sign bit is not 
> part of the payload. Requiring preservation of the sign bit means that one could 
> not negate a floating point value without checking for NaN, which would be a 
> significant performance penalty.
> 
> It's not a bug.


>From the Decimal Arithmetic Specification (http://speleotrove.com/decimal/decarith.pdf): 

"All special values may have a sign, as for finite numbers. The sign of an infinity is significant (that is, it is possible to have both positive and negative infinity), and the sign of a NaN has no meaning, although it may be considered part of the diagnostic information."

In addition, the specification, based on IEEE 754 and IEEE 854, makes a distinction between a "signaling NaN" (sNaN) and a "quiet NaN" (qNaN). D doesn't make this distinction. In essence all D NaNs are signaling.

The specification includes the quiet NaN for "a value representing undefined results... which does not cause an Invalid operation condition". In practice, this is the result of an arithmetic operation with a NaN (quiet or signaling) as an operand. (An sNaN as an operand will still raise an invalid operation flag, but if this doesn't cause an exception, the result of the operation will be a qNaN.)

I don't know if this will be of any more value than the NaN payload, but there it is.

Paul

p.s.