Fair points, but since `**` has its `Math.pow` counter part, why wouldn't `%%` have `Math.mod` as counterpart too? At least it looks like there's room for both, if standardized, as the behavior and description would likely be mostly the same (precedence a part)
On Thu, Aug 15, 2019 at 7:13 PM Isiah Meadows <[email protected]> wrote: > An operator is far more concise than a function call, and is likely to > see greater use. It also aligns better with peoples' intuition on what > the "modulus" is, avoiding subtle bugs like in `isOdd = x => x % 2 === > 1` (example from > https://en.wikipedia.org/wiki/Modulo_operation#Common_pitfalls - try > passing a negative to it). And given this one is high value (see > above) and *very* low cost (it can literally desugar to `(x % y + y) % > y`), I feel it does meet that bar. > > > 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). > > It has enough benefit I've seen CoffeeScript users default to `%%` and > only using `%` when they explicitly want the dividend-dependent > semantics. And engines with a native `%%`, if they can detect the > operands are always non-negative, can optimize it to `%` pretty > easily. It's better *enough* that you'd likely start seeing some > partially legitimate FUD spread about the standard `%`. > > One other added benefit of using divisor-dependent modulo is that `x > %% (2**n)`, where `x` and `n` are integers and `n >= 0`, could always > be safely rewritten to `x & (2**n - 1)` while still preserving > semantics, but `x % (2**n)` does *not* have this property. For > example: > > - `-1 %% (2**1)` → `-1 %% 1` → `1` > - `-1 & (2**1 - 1)` → `-1 & 1` → `1` > - `-1 % (2**1)` → `-1 % 2` → `-1` > > BTW, I literally tested all three of these in Chrome's devtools > console, using my `x %% y` → `(x % y + y) % y` desugaring. > > As for a native implementation and the spec, I'd recommend just doing > `copysign(fmod(x, y), y)` instead to retain precision. > > > 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.) > > I doubt this would happen with `%%`. It's similar enough to the > existing `%` in concept that most would expect it to have the same > precedence. With `**`, there was a very unique issue with it: there > were people actually *reading* it both ways, and even a language > (Python) that interprets `-a ** b` and `-a**b` *differently* in light > of that (as `(-a) ** b` and `-(a ** b)` respectively). That's not a > concern at all with most operators, so it doesn't apply to most new > operator proposals. > > ----- > > Isiah Meadows > [email protected] > www.isiahmeadows.com > On Thu, Aug 15, 2019 at 2:40 AM Claude Pache <[email protected]> > wrote: > > > > > > > > Le 12 août 2019 à 22:00, Matthew Morgan <[email protected]> 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 > > > > _______________________________________________ > > es-discuss mailing list > > [email protected] > > https://mail.mozilla.org/listinfo/es-discuss > _______________________________________________ > es-discuss mailing list > [email protected] > https://mail.mozilla.org/listinfo/es-discuss >
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