Re: percent-change

[Emanuel Berg]

Yuri Khan wrote:

>> But how/why does that work if 'from' is 0?
>>
>> (/ 1 0 0.01) ; 1.0e+INF
>>
>> What does that mean?
>
> In floating point, dividing by zero is allowed and yields
> +infinity, -infinity or (quiet) not-a-number depending on
> the sign of the numerator. ‘/’ knows to perform the division
> in floating when at least one operand is a float, which 0.01
> nicely provides. 1.0e+INF is just the printed representation
> for positive infinity.

A quiet NaN is not signalled. [1]

(/  1 0.0) ;  1.0e+INF
(/  0 0.0) ; -0.0e+NaN
(/ -1 0.0) ; -1.0e+INF

Yes, it is as you say, but why?

Maybe the answer is TLDR:
  https://en.wikipedia.org/wiki/Division_by_zero

[1] 
https://stackoverflow.com/questions/18118408/what-is-the-difference-between-quiet-nan-and-signaling-nan

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