How would the extension work in swift 3?

Question

I am using the below extension in swift 3, but I know it is incorrect because I still get errors. The point of the extension is to get a range of a string using the syntax string[0...2] That will return the string from 0. This is the error I am getting

extension String {

subscript (i: Int) -> Character {
    return self[self.characters.index(self.startIndex, offsetBy: i)]
}

subscript (i: Int) -> String {
    return String(self[i] as Character)
}

subscript (r: Range<Int>) -> String {
    let start = characters.index(startIndex, offsetBy: r.lowerBound)
    let end = characters.index(start, offsetBy: r.upperBound - r.lowerBound)
    return self[(start ..< end)]
  }
}

http://stackoverflow.com/a/38215613/2303865 – Leo Dabus Jul 18 '16 at 23:12 — Leo Dabus, Jul 18 '16 at 23:12

score 4 · Accepted Answer · answered Jul 18 '16 at 21:21

In Swift 3, the closed range operator ... generates CountableClosedRange for Ints. So you need to define one more overload for subscript.

    subscript (r: CountableClosedRange<Int>) -> String {
        let start = characters.index(startIndex, offsetBy: r.lowerBound)
        let end = characters.index(start, offsetBy: r.upperBound - r.lowerBound)
        return self[(start ... end)]
    }

score 0 · Answer 2 · answered Oct 28 '18 at 13:10

Updated in Swift 4:

extension String{
    subscript (r: CountableClosedRange<Int>) -> String {
        let start = index(startIndex, offsetBy: r.lowerBound)
        let end = index(start, offsetBy: r.upperBound - r.lowerBound)
        return String(self[start...end])
    }
}

How would the extension work in swift 3?

2 Answers2