On 14/04/2017 2:28 p.m., Bruno Le Floch wrote:
> On 04/02/2017 12:09 AM, Andrew Parsloe wrote:
>> Is there a simple way to distinguish 0 from 0 in l3fp?
>>
>> I had been aligning left in a table a column of sines, +0 or 0 followed
>> by decimal point & digits. By adding an \hphantom{} before the positive
>> numbers, they aligned nicely on the decimal point in the column, but
>> then along came 0. The simple conditional, "if !(number < 0) add
>> phantom", added the phantom and the column aligned like
>> 0
>> 0.5878
>> 0
>> 0.5878
>> etc.
>> In the absence of formatted printing of numbers, 0 is a problem. Even a
>> builtin test for 0 would ease matters, since in any kind of formatted
>> printing of numbers 0 and 0 will generally require different handling.
>>
>> As it is, I've had to test each fp for whether it is zero; if it is I
>> convert to a tl variable and test whether that is 0. This is clumsy.
>> Hence my opening question.
>>
>> Andrew
>
> Sorry for the delay.
>
> One option is to normalize 0 to +0 by adding +0. Namely, number+0 is
> equal to number for any nonzero number and is +0 for both +0 and 0.
>
> Another option is to compute 1/number after disabling the
> "divisionbyzero" trap, and test whether that's +inf or inf, but
> that's more complicated than the tl test you're doing.
>
> A better option is that I should add a "copysign" function, that is in
> the IEEE standard: copysign(1,x) gives +1 or 1 depending on the sign
> bit of x (so in particular this distinguishes +0 and 0).
>
> For your specific use case I would use \fp_to_tl:n then test whether the
> first character (\str_head:n) is "" or not. This also covers nan properly.
>
> Bruno
>
Ah, the "add +0" suggestion is helpful, and certainly simpler than what
I was doing.
You implemented, very promptly, an earlier suggestion of mine about
adding "sign" to l3fp. Thank you. In the course of dealing with the
above problem I noticed that sign(0) = 0. I presume this is the
expected result. (I was rather hoping sign(x) might evaluate to one of
the *three* values 1, 0, 1.)
Andrew

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