[swift-evolution] Feature proposal: Range operator with step

[Erica Sadun]

> On Apr 6, 2016, at 3:05 PM, Dave Abrahams via swift-evolution <swift-evolution at swift.org> wrote:
> 
> 
> on Wed Apr 06 2016, Xiaodi Wu <xiaodi.wu-AT-gmail.com <http://xiaodi.wu-at-gmail.com/>> wrote:
> 
>> On Wed, Apr 6, 2016 at 3:28 PM, Dave Abrahams via swift-evolution
>> <swift-evolution at swift.org> wrote:
>>> You if you need to represent `<..` intervals in scientific computing,
>>> that's a pretty compelling argument for supporting them.
>>> 
>>>> I'd like to be able to represent any of those as
>>>> Intervals-which-are-now-Ranges. It makes sense to do so because the
>>>> things I want to do with them, such as clamping and testing if some
>>>> value is contained, are exactly what Intervals-now-Ranges provide.
>>>> Looking around, it seems many other languages provide only what Swift
>>>> currently does, but Perl does provide `..`, `..^`, `^..`, and `^..^`
>>>> (which, brought over to Swift, would be `...`, `..<`, `<..`, and
>>>> `<.<`).
>>> 
>>> Do we need fully-open ranges too?
>> 
>> I haven't encountered a need for open ranges, but I would expect that
>> other applications in scientific computing could make use of them.
>> I rather like Pyry's suggestions below. 
> 
> Below?

Logically in time below.

I believe the following is a valid conversion of the Xiaodi Wu below into the Dave A domain.

> On Apr 6, 2016, at 2:29 PM, Pyry Jahkola via swift-evolution <swift-evolution at swift.org> wrote:
> 
> I think a sensible specification would be that with a positive step size, the count starts from the lower bound, and with a negative one, it starts from the upper bound (inclusive or exclusive). Thus, the following examples should cover all the corner cases:
> 
>     (0 ... 9).striding(by: 2) == [0, 2, 4, 6, 8]
>     (0 ..< 9).striding(by: 2) == [0, 2, 4, 6, 8]
>     (0 <.. 9).striding(by: 2) ==    [2, 4, 6, 8]
>     (0 <.< 9).striding(by: 2) ==    [2, 4, 6, 8]
> 
>     (0 ... 9).striding(by: 3) == [0, 3, 6, 9]
>     (0 ..< 9).striding(by: 3) == [0, 3, 6]
>     (0 <.. 9).striding(by: 3) ==    [3, 6, 9]
>     (0 <.< 9).striding(by: 3) ==    [3, 6]
> 
>     (0 ... 9).striding(by: -2) == [9, 7, 5, 3, 1]
>     (0 ..< 9).striding(by: -2) ==    [7, 5, 3, 1]
>     (0 <.. 9).striding(by: -2) == [9, 7, 5, 3, 1]
>     (0 <.< 9).striding(by: -2) ==    [7, 5, 3, 1]
> 
>     (0 ... 9).striding(by: -3) == [9, 6, 3, 0]
>     (0 ..< 9).striding(by: -3) ==    [6, 3, 0]
>     (0 <.. 9).striding(by: -3) == [9, 6, 3]
>     (0 <.< 9).striding(by: -3) ==    [6, 3]
> 
> Lastly, if you want the positive stride reversed, you'd do just that:
> 
>     (0 ... 9).striding(by: 2).reverse() == [8, 6, 4, 2, 0]
> 
> — Pyry

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