Get nth character of a string in Swift programming language

Question

How can I get the nth character of a string? I tried bracket([]) accessor with no luck.

var string = \"Hello, world!\"

var firstChar = string[         


        

Answer 1

My very simple solution:

Swift 4.1:

let myString = "Test string"
let index = 0
let firstCharacter = myString[String.Index(encodedOffset: index)]

Swift 5.1:

let firstCharacter = myString[String.Index.init(utf16Offset: index, in: myString)]

Answer 2

No indexing using integers, only using String.Index. Mostly with linear complexity. You can also create ranges from String.Index and get substrings using them.

Swift 3.0

let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.index(before: someString.endIndex)]
let charAtIndex = someString[someString.index(someString.startIndex, offsetBy: 10)]

let range = someString.startIndex..<someString.index(someString.startIndex, offsetBy: 10)
let substring = someString[range]

Swift 2.x

let firstChar = someString[someString.startIndex]
let lastChar = someString[someString.endIndex.predecessor()]
let charAtIndex = someString[someString.startIndex.advanceBy(10)]

let range = someString.startIndex..<someString.startIndex.advanceBy(10)
let subtring = someString[range]

Note that you can't ever use an index (or range) created from one string to another string

let index10 = someString.startIndex.advanceBy(10)

//will compile
//sometimes it will work but sometimes it will crash or result in undefined behaviour
let charFromAnotherString = anotherString[index10]

Answer 3

Swift 4

let str = "My String"

String at index

let index = str.index(str.startIndex, offsetBy: 3)
String(str[index])    // "S"

Substring

let startIndex = str.index(str.startIndex, offsetBy: 3)
let endIndex = str.index(str.startIndex, offsetBy: 7)
String(str[startIndex...endIndex])     // "Strin"

First n chars

let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[..<startIndex])    // "My "

Last n chars

let startIndex = str.index(str.startIndex, offsetBy: 3)
String(str[startIndex...])    // "String"

---

Swift 2 and 3

str = "My String"

**String At Index **

Swift 2

let charAtIndex = String(str[str.startIndex.advancedBy(3)])  // charAtIndex = "S"

Swift 3

str[str.index(str.startIndex, offsetBy: 3)]

SubString fromIndex toIndex

Swift 2

let subStr = str[str.startIndex.advancedBy(3)...str.startIndex.advancedBy(7)] // subStr = "Strin"

Swift 3

str[str.index(str.startIndex, offsetBy: 3)...str.index(str.startIndex, offsetBy: 7)]

First n chars

let first2Chars = String(str.characters.prefix(2)) // first2Chars = "My"

Last n chars

let last3Chars = String(str.characters.suffix(3)) // last3Chars = "ing"

Answer 4

As an aside note, there are a few functions applyable directly to the Character-chain representation of a String, like this:

var string = "Hello, playground"
let firstCharacter = string.characters.first // returns "H"
let lastCharacter = string.characters.last // returns "d"

The result is of type Character, but you can cast it to a String.

Or this:

let reversedString = String(string.characters.reverse())
// returns "dnuorgyalp ,olleH" 

:-)

Answer 5

I just came up with this neat workaround

var firstChar = Array(string)[0]

Answer 6

My solution is in one line, supposing cadena is the string and 4 is the nth position that you want:

let character = cadena[advance(cadena.startIndex, 4)]

Simple... I suppose Swift will include more things about substrings in future versions.