[Cfp-interest 2846] Re: The leading (or negative field) bit on a NaN

[Steve (Swift) Canon]

This is one of the dustier corners of 754; the relevant part of the standard is:

> When either an input or result is NaN, this standard does not interpret the sign of a NaN. Note, however, that operations on bit strings — copy, negate, abs, copySign — specify the sign bit of a NaN result, sometimes based upon the sign bit of a NaN operand.

So NaN has a signbit in the encoding, but it is not a sign, but it behaves as though it were a sign for the purposes of the copySign operation.

Note that the signbit is explicitly _not_ part of a NaN’s payload, even though Mike finds it convenient to work with them this way:

Binary formats:
> The remaining bits, which are in the trailing significand field, encode the payload, which might be diagnostic information (see above).

Decimal:
> The remaining bits of G are ignored, and T constitutes the NaN’s payload, which can be used to distinguish various NaNs.

– Steve

> On Aug 24, 2023, at 10:43 AM, Mike Cowlishaw <mfc at speleotrove.com> wrote:
> 
> I would say that's a question to ask on the 754 list, not here :-).  By
> definition NaNs are not numbers so do not have a sign.  I just find it
> convenient to think of the NaN's payload as having a sign -- which is not
> suggesting that the NaN has a sign.
> 
> Mike 
> 
>> -----Original Message-----
>> On Thu, 24 Aug 2023, Mike Cowlishaw wrote:
>> 
>>> Technically its 'just another bit'.  However, in my 
>>> testcases I treat 
>>> the payload of the NaN as being a signed integer having the 
>>> value of 
>>> the significand and signed by treating that bit as a sign 
>>> bit.  This 
>>> to me, is useful and unambiguous, and it also works for 
>>> both decimal 
>>> significand codings..
>> 
>> I was looking at in the context of the projection of a complex number
>> 
>> x + I y
>> 
>> onto the eextended complex plane (as modelled by a Riemann 
>> sphere) of a complex number with an infinite part. This is 
>> the 'cproj' routine.
>> 
>> It returns
>> 
>> INFINOYI + I (+|-0.0)
>> 
>> where the sign of that zero is directly affected by the sign of 'y'.
>> 
>> However, I cannot see how, if 'y' is a NaN, that its payload 
>> details, specifically the 'sign' bit, should influence the 
>> sign of that 0.0.
>> 
>> Just curious.
>> 
>> Or is the reason definitional to provide reproducible results?
>> 
>> Thanks - Damian
>> 
> 
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