Hi there!

Ole Begeman offers here <https://oleb.net/blog/2016/09/swift-3-ranges/> (take a 
look at the bottom of the page) an interesting consideration about converting 
between half-open and closed ranges.

As of now, it seems the way to go is by overloading…


import Foundation

func random(from range: Range<Int>) -> Int {
    let lowerBound = range.lowerBound
    let upperBound = range.upperBound
    
    return lowerBound + Int(arc4random_uniform(UInt32(upperBound - lowerBound)))
}

func random(from range: ClosedRange<Int>) -> Int {
    let lowerBound = range.lowerBound
    let upperBound = range.upperBound
    
    return lowerBound + Int(arc4random_uniform(UInt32(upperBound - lowerBound + 
1)))
}

let r1 = random(from: 4 ..< 8)
let r2 = random(from: 6 ... 8)


Cheers,

— A

> On Oct 12, 2016, at 6:21 AM, Jean-Denis Muys via swift-users 
> <swift-users@swift.org> wrote:
> 
> Hi,
> 
> I defined this:
> 
> func random(from r: Range<Int>) -> Int {
>     let from = r.lowerBound
>     let to =  r.upperBound
>     
>     let rnd = arc4random_uniform(UInt32(to-from))
>     return from + Int(rnd)
> }
> 
> so that I can do:
> 
> let testRandomValue = random(from: 4..<8)
> 
> But this will not let me do:
> 
> let otherTestRandomValue = random(from: 4...10)
> 
> The error message is a bit cryptic:
> 
> “No ‘…’ candidate produce the expected contextual result type ‘Range<Int>’”
> 
> What is happening is that 4…10 is not a Range, but a ClosedRange.
> 
> Of course I can overload my function above to add a version that takes a 
> ClosedRange.
> 
> But this is not very DRY.
> 
> What would be a more idiomatic way?
> 
> Thanks,
> 
> Jean-Denis
> 
> _______________________________________________
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> swift-users@swift.org
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