On Mon, Mar 14, 2016 at 2:16 PM, Tab Atkins Jr. <jackalm...@gmail.com> wrote:
> On Fri, Mar 11, 2016 at 8:17 PM, Isiah Meadows <isiahmead...@gmail.com> > wrote: > > In my honest opinion, there's not much reason to just require the sort > to be > > stable. Some engines have done this in the past, and the spec technically > > allows it. At this point, stable sorts are about as fast as unstable > ones, > > both in theory and practice (wasn't the case 10 years ago IIRC). > > I think you meant "to not just require", yes? As in, you think the > spec *should* require .sort() to be stable? > > On Sun, Mar 13, 2016 at 12:23 PM, Vic99999 <vic99...@yandex.ru> wrote: > > Stable sort is useful sometimes, but it is possible to implement it in > js. > > Most things in the JS standard library are possible in userland code, > so that's not a useful rebuttal to anything by itself. The question > is whether it's worth requiring that the standard sort be stable. > Stable sort is never *bad* - there are no use-cases where a stable > sort gives the wrong answer while an unstable sort is correct, but the > opposite is definitely true. The only question is how much > slower/more expensiver stable sorting is than unstable sorting, and > whether that cost is sufficient to outweigh the usefulness of a stable > sort. > ISTM a symbol could be defined to grab a handle on the implementation's fastest possible sort logic. If the implementation has a substantially faster unstable sort algo they'd surface it here -- otherwise this would just delegate to the implementation's stable sort. This could be `Array.prototype.sort`, or it could also be a spec'd symbol which `Array.prototype.sort` delegates to. (Or this could all be done w/ modules, although modules might complicate what's otherwise pretty simple to polyfill). But fastest-possible (likely unstable) sort logic should be opt-in, not opt-out. If the perf difference is that relevant to your use case you probably don't want off-the-shelf `Array.prototype.sort` anyway.
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