Modulo Operator %%
Andrea Giammarchi
andrea.giammarchi at gmail.com
Thu Aug 15 07:46:57 UTC 2019
FWIW another disadvantage is that operators cannot be polyfilled, so it'll
take forever for those not using transpilers to adopt these, while having a
`Math,mod` would work right away
On Thu, Aug 15, 2019 at 8:40 AM Claude Pache <claude.pache at gmail.com> wrote:
>
>
> Le 12 août 2019 à 22:00, Matthew Morgan <mmm211 at zips.uakron.edu> a écrit :
>
> JS needs a modulo operator. It currently has the remainder operator `%`
> which works in most cases except for negative values. I believe the the
> `%%` would work great and be easy to remember.
>
> let x = (-13) %% 64;
> is equivalent to
> let x = ((-13 % 64) + 64) % 64;
>
>
> Is there a strong advantage of an `%%` operator over a `Math.mod()`
> function? There is the precedent of the `**` operator implemented as
> alternative of `Math.pow()` few years ago. It would be interesting to hear
> the feedback of those that use regularly powers, whether the benefit was
> clear (personally, I almost never use either `Math.pow()` or `**`, so that
> I can’t say anything).
>
> At least one disadvantage of an operator over a function, is that you have
> to think about precedence. The problem is exacerbated in JS, because
> (following some other languages) the unary minus has an uncanny high
> precedence level, confusingly very different than the one of the binary
> minus; so that, after having designed `**`, it was realised at the last
> minute that `-a**b` would be dumbly interpreted as `(-a)**b` instead of
> `-(a**b)` or `0-a**b`, as anybody who would be likely to actually use the
> operator would expect. (That particular issue was resolved in a hurry by
> making the parenthesis-left form a syntax error.)
>
> —Claude
>
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