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.
Oh! In my application, time flows downward.
I believe the following is a valid conversion of the Xiaodi Wu below into the
Dave A domain.
    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]
These all look reasonable to me.
    Lastly, if you want the positive stride reversed, you'd do just that:
    (0 ... 9).striding(by: 2).reverse() == [8, 6, 4, 2, 0]
Also reasonable.
···
on Wed Apr 06 2016, Erica Sadun <erica-AT-ericasadun.com> wrote:
    On Apr 6, 2016, at 3:05 PM, Dave Abrahams via swift-evolution >     <swift-evolution@swift.org> wrote:
    on Wed Apr 06 2016, Xiaodi Wu <xiaodi.wu-AT-gmail.com> wrote:
        On Wed, Apr 6, 2016 at 3:28 PM, Dave Abrahams via swift-evolution >         <swift-evolution@swift.org> wrote:
    On Apr 6, 2016, at 2:29 PM, Pyry Jahkola via swift-evolution >     <swift-evolution@swift.org> wrote:
--
Dave