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

[davesweeris at mac.com]

It’s not a matter of floating point error accumulation… At least on my machine, once a Double hits +/-∞, there’s no way that I know of to get back to normal floating point numbers. That is to say, for *all* normal, finite values of x, "-Double.infinity + x" will just return “-inf". If x is to equal Double.infinity, Double.NaN, or Double.quietNaN, then it’ll return “nan” (which, incidentally, will fail the regular equality test… Double.NaN isn’t even equal to itself; I think checking the floating point class is the way to do it).

I could easily be missing something, but AFAICT the only way to always get the correct sequence (without splitting the floating point types off into their own thing) is either have a negative stride swap start and end *before* the StrideTo starts generating values (that is, *not* by just calling `.reverse()` on something with a positive stride), or to allow “0 ..< -Double.infinity” to be a valid range (with the negative stride being implied).

- Dave Sweeris

> On Apr 9, 2016, at 6:59 PM, Xiaodi Wu <xiaodi.wu at gmail.com> wrote:
> 
> Yikes. Not too concerned about the infinite loop issue, as floating point strides when fixed to avoid error accumulation will necessarily enforce a finite number of steps. However, you're talking a regular, not-at-all-lazy Array being returned? That would be not good at all...
> 
> On Sun, Apr 10, 2016 at 12:29 AM Dave via swift-evolution <swift-evolution at swift.org <mailto:swift-evolution at swift.org>> wrote:
> 
>> On Apr 9, 2016, at 4:33 AM, Haravikk via swift-evolution <swift-evolution at swift.org <mailto: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<>, 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
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