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How to convert "Index" to type "Int" in Swift?

Tags:

string

swift

edit/update:

Xcode 11 • Swift 5.1 or later

extension StringProtocol {
    func distance(of element: Element) -> Int? { firstIndex(of: element)?.distance(in: self) }
    func distance<S: StringProtocol>(of string: S) -> Int? { range(of: string)?.lowerBound.distance(in: self) }
}

extension Collection {
    func distance(to index: Index) -> Int { distance(from: startIndex, to: index) }
}

extension String.Index {
    func distance<S: StringProtocol>(in string: S) -> Int { string.distance(to: self) }
}

Playground testing

let letters = "abcdefg"

let char: Character = "c"
if let distance = letters.distance(of: char) {
    print("character \(char) was found at position #\(distance)")   // "character c was found at position #2\n"
} else {
    print("character \(char) was not found")
}

let string = "cde"
if let distance = letters.distance(of: string) {
    print("string \(string) was found at position #\(distance)")   // "string cde was found at position #2\n"
} else {
    print("string \(string) was not found")
}

Works for Xcode 13 and Swift 5

let myString = "Hello World"

if let i = myString.firstIndex(of: "o") {
  let index: Int = myString.distance(from: myString.startIndex, to: i)
  print(index) // Prints 4
}

The function func distance(from start: String.Index, to end: String.Index) -> String.IndexDistance returns an IndexDistance which is just a typealias for Int


Swift 4

var str = "abcdefg"
let index = str.index(of: "c")?.encodedOffset // Result: 2

Note: If String contains same multiple characters, it will just get the nearest one from left

var str = "abcdefgc"
let index = str.index(of: "c")?.encodedOffset // Result: 2

encodedOffset has deprecated from Swift 4.2.

Deprecation message: encodedOffset has been deprecated as most common usage is incorrect. Use utf16Offset(in:) to achieve the same behavior.

So we can use utf16Offset(in:) like this:

var str = "abcdefgc"
let index = str.index(of: "c")?.utf16Offset(in: str) // Result: 2