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

Dave Abrahams dabrahams at apple.com
Sat Apr 9 20:58:52 CDT 2016

```on Sat Apr 09 2016, davesweeris-AT-mac.com 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<>,

.reversed() returns a ReversedCollection when the underlying collection
is bidirectional:

https://github.com/apple/swift/blob/swift-3-indexing-model/stdlib/public/core/Reverse.swift#L250

That's lazy and cheap.

> 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
>

--
Dave
```