What is the nil or "empty" countable range in Swift?

Question

In Swift3,

let highestIndex = 7
for index in 1 ..< highestIndex {
    latter(index)
}

however,

let highestIndex = 0, or anything smaller
for index in 1 ..< highestIndex {
    latter(index)
}

that crashes.

So, the ..< operator is impractical in most situations.

So, perhaps, something like this ...

for index in (1 ..< highestIndex).safely() {
    latter(index)
}

How would one write the extension safely ?

Or at worst just a function safely(from,to) which returns a CountableRange, or, an "empty" (?) countable range? (I don't understand if a Swift extension can "catch" an error?)

"*Is there such a thing as an "empty" countable range*" – yes, `1..<1` is just that (it does not crash). — Hamish, Feb 05 '17 at 22:49
If you don't want the loop to be entered if the upper bound is less than the lower bound, `stride` is a good alternative – see for example http://stackoverflow.com/a/36861439/2976878 — Hamish, Feb 05 '17 at 22:52
[DOH facepalm](http://cdn.onegreenplanet.org/wp-content/uploads/2010/10/2014/11/lead1.jpg) -- **stride**, of course, that's the way to go. you've cracked it again, @Hamish — Fattie, Feb 05 '17 at 22:55

score 2 · Answer 1 · answered Feb 05 '17 at 22:47

Sticking as closely to the ..< operator as possible the following would be an appropriate implementation of safely.

public func safely<Bound>(_ minimum: Bound, _ maximum: Bound) -> CountableRange<Bound> where Bound : _Strideable & Comparable, Bound.Stride : Integer {
    if maximum < minimum {
        return CountableRange(uncheckedBounds: (lower: minimum, upper: minimum))
    }
    return CountableRange(uncheckedBounds: (lower: minimum, upper: maximum))
}

let one = safely(0, 12)  // CountableRange(0..<12)
let two = safely(15, 12) // CountableRange(15..<15)

one.isEmpty // false
two.isEmpty // true

You might want to consider giving the parameters better names to indicate wether or not they are inclusive / exclusive bounds, etc.

Creating an extension and adding a .safely() will not work since the bounds are checked when creating the CountableRange before the call to safely could even be done.

Say, in your excellent code. I'd suggest, the first return should just be "return 0..<0". Emphasizing the fact that you're just returning "no trips". Great stuff. — Fattie, Feb 05 '17 at 22:56

score 1 · Answer 2 · answered Feb 05 '17 at 22:41

1

I'm able to get this to work fine:

for index in 1 ..< 1 {
    latter(index)
}

For the <1 case, I recommend something like this:

guard highestIndex > 0 else { return }

for index in 1 ..< highestIndex {
    latter(index)
}

answered Feb 05 '17 at 22:41

Mark

7,167
4
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Hey Mark - thanks for that. Sure, "guard" is a terrific feature. But, you know, it's no good for inline code; the most natural situation is that of course, you just want it to "not do any iterations" if the value is too low.... – Fattie Feb 05 '17 at 22:47

score 1 · Accepted Answer · answered Feb 06 '17 at 11:16

1

I am a fan of simple solutions:

let lowerIndex = 1
let higherIndex = 0
for index in lowerIndex ..< max(lowerIndex, higherIndex) {
    latter(index)
}

answered Feb 06 '17 at 11:16

Sulthan

128,090
22
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270

agreed - KISS. I guess, one could just put that in a custom infix operator, and that's it. – Fattie Feb 06 '17 at 11:24

What is the nil or "empty" countable range in Swift?

3 Answers3