Swift: convert range to array

protocol ArrayRepresentable {
    typealias ArrayType
    
    func toArray() -> ArrayType[]
}

extension Range : ArrayRepresentable {
    func toArray() -> T[] {
        return T[](self)
    }
}



(1..5).toArray() // => [1, 2, 3, 4]

(-2.0..2.0).toArray() // => [-2.0, -1.0, 0.0, 1.0]

Swift: generic function to convert a sequence to an array

func toArray<S : Sequence>(seq: S) -> Array<S.GeneratorType.Element> {
    return Array<S.GeneratorType.Element>(seq)
}


toArray(1..5) // => [1, 2, 3, 4]

toArray(-2.0..2.0) // => [-2.0, -1.0, 0.0, 1.0]

@andy318. Great stuff, thanks a lot! Works great.

What does this provide above Array(1...42)?

Swift 3?

👍

@eonist
(1...40).map { String($0) }
Swift 3

Meta: I googled this again. And found my self asking the same question. :P

Thanks @RomanVolkov was just about to post that same thing.

I found this to be the best option in Swift 3:

Array(0...3)    // [0,1,2,3]
Array(0..<3)    // [0,1,2]

import Foundation

public extension Range where Bound: Strideable, Bound.Stride: SignedInteger {

  /// Convert to an array.
  func asArray() -> [Bound] {
    Array(self)
  }
}

public extension ClosedRange where Bound: Strideable, Bound.Stride: SignedInteger {

  /// Convert to an array.
  func asArray() -> [Bound] {
    Array(self)
  }
}

public extension Sequence {

  /// Convert to an array.
  func asArray() -> [Iterator.Element] {
    Array(self)
  }
}

(1 ..< 4).asArray() // [1, 2, 3]
(1 ... 4).asArray() // [1, 2, 3, 4]
zip([1, 2, 3], ["a", "b", "c"]).asArray() // [(1, "a"), (2, "b"), (3, "c")]