[swift-evolution] [Draft]: Introducing a striding(by:) method on 3.0 ranges

[Xiaodi Wu]

Yikes. Not too concerned about the infinite loop issue, as floating point
strides when fixed to avoid error accumulation will necessarily enforce a
finite number of steps. However, you're talking a regular, not-at-all-lazy
Array being returned? That would be not good at all...

On Sun, Apr 10, 2016 at 12:29 AM Dave via swift-evolution <
swift-evolution at swift.org> wrote:

>
> On Apr 9, 2016, at 4:33 AM, Haravikk via swift-evolution <
> swift-evolution at swift.org> wrote:
>
> While I’m in favour of the basic idea I think the operator selection is
> too complex, and I’m not sure about the need for negative strides. Really
> all I want are the following:
>
> (0 ... 6).striding(by: 2) // [0, 2, 4, 6] x from 0 to 6
> (0 ..< 6).striding(by: 2) // [0, 2, 4] x from 0 while <6
> (6 ... 0).striding(by: 2) // [6, 4, 2, 0] x from 6 to 0
> (6 ..> 0).striding(by: 2) // [6, 4, 2] x from 6 while >0
>
> Everything else should be coverable either by flipping the order, or using
> .reverse(). The main advantage is that there’s only one new operator to
> clarify the 6 ..> 0 case, though you could always just reuse the existing
> operator if you just interpret it as “x from 6 to, but not including, 0"
>
>
> `.reverse()` returns an array, though, not a StrideTo<>, which means it’ll
> get in an infinite loop on infinite sequences. This works fine:
> for i in stride(from: 0.0, to: Double.infinity, by: M_PI) {
>     if someTestInvolving(i) { break }
>     ...
> }
>
> But this never even starts executing the loop because of the infinite loop
> inside `.reverse()`:
> for i in stride(from: -Double.infinity, to: 0.0, by: M_PI).reverse() {
>     if someTestInvolving(i) { break }
>     ...
> }
>
> - Dave Sweeris
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