How do I make this extension of Array?

Asked Jan 20 '15 at 10:40

Active Jan 20 '15 at 10:40

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I am trying to extend the Array type with a computed property called randElement, that's, obviously, returning a random element of the array, but I don't know how to deal with the fact that the type of elements that the Array contains.
Here's what I'm trying to do:

extension Array {
    var randElement:Array.Type {
        var randNumber = Double(arc4random())/Double(UInt32.max)
        randNumber*=(Double(self.count-1))
        randNumber = floor(randNumber)
        return self[Int(randNumber)]
    }
}

It triggers me this error at the level of the return:
'T' is not convertible to 'Array<T>.Type

What am I doing wrong?

asked Jan 20 '15 at 10:40

Pop Flamingo

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This answer should be what you are looking for: http://stackoverflow.com/a/24101606/1187415. – Martin R Jan 20 '15 at 10:45
@MartinR I want to use a computed property rather than a function... – Pop Flamingo Jan 20 '15 at 10:48
How, well, that's OK actually. – Pop Flamingo Jan 20 '15 at 10:48
The main point is that `Array` is defined as `struct Array` where `T` is the element type, so your function or property or whatever must return `T`. – Martin R Jan 20 '15 at 10:50
Note also that `arc4random_uniform()` is generally preferred over `arc4random()`. – Martin R Jan 20 '15 at 10:53
@MartinR How thank you for your information about `struct Array`, know I understand why I can directly use T. And about `arc4random_uniform()`, I'm now using it based on what I saw on the other post. – Pop Flamingo Jan 20 '15 at 10:55

How do I make this extension of Array?

0 Answers0