How do you split a String into a [String] and not [Substring]?

Question

Consider the following extension:

extension Array where Element == String {
    func foo(){
    }
}

And consider the following string that I want to split so I can use the extension...

let string = "This|is|a|test"
let words = string.split(separator: "|")

The problem is words is a [Substring] and not [String] so I can't call foo() on it.

So in Swift 4, how do I split a string to return a [String] and not a [Substring]?

string.components(separatedBy: .whitespaces) will return an array of strings — Leo Dabus
– Leo Dabus, Commented Mar 10, 2018 at 6:43
If you really need to use split method you can map them into strings .map(String.init) — Leo Dabus
– Leo Dabus, Commented Mar 10, 2018 at 6:45
Ideally you'd modify the extension to work with substrings, I presume — John Dvorak
– John Dvorak, Commented Mar 10, 2018 at 6:46
@JohnDvorak, I had just posted a second question specifically asking how to write an extension that works with both. Can you elaborate in an answer? — Mark A. Donohoe
– Mark A. Donohoe, Commented Mar 10, 2018 at 6:47

Martin R · Accepted Answer · 2018-03-10 07:27:13Z

14

As Leo said above, you can use components(separatedBy:)

let string = "This|is|a|test"
let words = string.components(separatedBy: "|")
words.foo()

instead, that returns a [String].

If you want to stick with split() (e.g. because it has more options, such as to omit empty subsequences), then you'll have to create a new array by converting each Substring to a String:

let string = "This|is|a|test"
let words = string.split(separator: "|").map(String.init)
words.foo()

Alternatively – if possible – make the array extension method more general to take arguments conforming to the StringProtocol protocol, that covers both String and Substring:

extension Array where Element: StringProtocol {
    func foo(){
    }
}

edited Mar 10, 2018 at 7:27

answered Mar 10, 2018 at 6:46

Martin R

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1 Comment

Mark A. Donohoe Over a year ago

This works perfectly. And your StringProtocol solution actually answers my other question that I just posted. Feel free to copy it there and I'll accept that there as well.

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