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
