On Wed, Apr 6, 2016 at 1:29 PM Pyry Jahkola via swift-evolution < swift-evolution@swift.org> wrote:
> > On 06 Apr 2016, at 23:17, Dave Abrahams via swift-evolution < > swift-evolution@swift.org> wrote: > > 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 > > > 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] > I have always desired for a complete set of range variants like you outlined above. I don't have current examples warranting a complete set but I could easily see having to work on a data set that may be best done by having a complete set of range variants so you don't need to transform the data set to get it to work with the language you happen to be using. It also better supports floating point ranges which may not have easy ways to transform the data. I also like the suggestion of how striding and reverse should work. -Shawn
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