Bug in Number.prototype.toFixed for 0.5 and -0.5

[Sundararajan Athijegannathan]

Yes, your interpretation seems right and I confirm that nashorn has a bug.

https://es5.github.io/#x15.7.4.5

Else, /x/ < 10^21

 1. Let /n/ be an integer for which the exact mathematical value of /n/
    ÷ 10^f – /x/ is as close to zero as possible. If there are two such
    /n/, pick the larger /n/.

I filed https://bugs.openjdk.java.net/browse/JDK-8135000

Thanks for reporting this issue.
-Sundar


On 9/3/2015 3:29 AM, Esben Andreasen wrote:
> The implementation of Number.prototype.toFixed seems to be wrong in
> two cases. Can you confirm this, and/or report it on an issue tracker?
>
> Observed behaviour:
>
>> $ jjs -version
>> nashorn 1.8.0_60
>> jjs> 0.5.toFixed()
>> 0
>> jjs> -0.5.toFixed()
>> 0
> Expected behaviour:
>
> The two results should be 1 and -1 respectively.
>
> I expect this behaviour because that is how I interpret ES5 and ES6
> specifications for Number.prototype.toFixed, and because that is the
> behaviour that I can observe in Chrome and Firefox.
>
>
>
> Extra comments on possible cause of bug:
>
> The interesting part of the ES6 specification is
>
> 20.1.3.3 - 10. - a.:
>
>> Let n be an integer for which the exact mathematical value of n /
>    10^f – x is as close to zero as possible. If there are two such n,
>    pick the larger n.
>
> In the case of `0.5.toFixed()` the expression above becomes:
>
>> n / 10^0 - 0.5 = n / 1 - 0.5 = n - 0.5
> which solves to `n = 0` or `n = 1`.  It seems that Nashorn picks the
> `n = 0` case instead of the `n = 1` case.
>
>
>
> ---
> Esben Andreasen