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

[Dave Abrahams]

on Wed Apr 06 2016, Erica Sadun <swift-evolution at swift.org> wrote:

>> On Apr 6, 2016, at 12:43 PM, Dave Abrahams <dabrahams at apple.com> wrote:
>> 
>> 
>> on Wed Apr 06 2016, Erica Sadun <erica-AT-ericasadun.com> wrote:
>> 
>
>>>    On Apr 6, 2016, at 12:16 PM, Dave Abrahams via swift-evolution
>>>    <swift-evolution at swift.org> wrote:
>>>    (0..<199).striding(by: -2)
>>> 
>>>    are even or odd.
>>> 
>>> (0..<199).striding(by: -2): 0..<199 == 0...198 Even
>>> (1..<199).striding(by: -2): 1..<199 == 1...198 Even
>> 
>> I understand the logic that got you there, but I find it incredibly
>> counter-intuitive that striding by 2s over a range with odd endpoints
>> should produce even numbers... I can't imagine any way I'd be convinced
>> that was a good idea.
>
> Guidance:
>
> When using odd integer literals to produce an even number sequence,
> prefer the `...` operator to the `..<` operator and change your ending
> literal to an even number.

I don't think you can fix counterintuitive behavior with guidance.  

(1..<199).striding(by: -2) is the first way I'd reach for to express
197, 195, ..., 3, 1

>
>
> -- E
>
>> 
>>> (0..<198).striding(by: -2): 1..<198 == 0...197 Odd
>>> (1..<198).striding(by: -2): 1..<198 == 1...197 Odd
>>> 
>>> -- E
>>> 
>> 
>> -- 
>> Dave
>
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-- 
Dave