Swift 5.2

let str = "abcdef"
str[1 ..< 3] // returns "bc"
str[5] // returns "f"
str[80] // returns ""
str.substring(fromIndex: 3) // returns "def"
str.substring(toIndex: str.length - 2) // returns "abcd"

You will need to add this String extension to your project (it's fully tested):

extension String {

    var length: Int {
        return count
    }

    subscript (i: Int) -> String {
        return self[i ..< i + 1]
    }

    func substring(fromIndex: Int) -> String {
        return self[min(fromIndex, length) ..< length]
    }

    func substring(toIndex: Int) -> String {
        return self[0 ..< max(0, toIndex)]
    }

    subscript (r: Range<Int>) -> String {
        let range = Range(uncheckedBounds: (lower: max(0, min(length, r.lowerBound)),
                                            upper: min(length, max(0, r.upperBound))))
        let start = index(startIndex, offsetBy: range.lowerBound)
        let end = index(start, offsetBy: range.upperBound - range.lowerBound)
        return String(self[start ..< end])
    }
}

Even though Swift always had out of the box solution to this problem (without String extension, which I provided below), I still would strongly recommend using the extension. Why? Because it saved me tens of hours of painful migration from early versions of Swift, where String's syntax was changing almost every release, but all I needed to do was to update the extension's implementation as opposed to refactoring the entire project. Make your choice.

let str = "Hello, world!"
let index = str.index(str.startIndex, offsetBy: 4)
str[index] // returns Character 'o'

let endIndex = str.index(str.endIndex, offsetBy:-2)
str[index ..< endIndex] // returns String "o, worl"

String(str.suffix(from: index)) // returns String "o, world!"
String(str.prefix(upTo: index)) // returns String "Hell"

I just came up with this neat workaround

var firstChar = Array(string)[0]

Xcode 11 • Swift 5.1

You can extend StringProtocol to make the subscript available also to the substrings:

extension StringProtocol {
    subscript(_ offset: Int)                     -> Element     { self[index(startIndex, offsetBy: offset)] }
    subscript(_ range: Range<Int>)               -> SubSequence { prefix(range.lowerBound+range.count).suffix(range.count) }
    subscript(_ range: ClosedRange<Int>)         -> SubSequence { prefix(range.lowerBound+range.count).suffix(range.count) }
    subscript(_ range: PartialRangeThrough<Int>) -> SubSequence { prefix(range.upperBound.advanced(by: 1)) }
    subscript(_ range: PartialRangeUpTo<Int>)    -> SubSequence { prefix(range.upperBound) }
    subscript(_ range: PartialRangeFrom<Int>)    -> SubSequence { suffix(Swift.max(0, count-range.lowerBound)) }
}

extension LosslessStringConvertible {
    var string: String { .init(self) }
}

extension BidirectionalCollection {
    subscript(safe offset: Int) -> Element? {
        guard !isEmpty, let i = index(startIndex, offsetBy: offset, limitedBy: index(before: endIndex)) else { return nil }
        return self[i]
    }
}

Testing

let test = "Hello USA 🇺🇸!!! Hello Brazil 🇧🇷!!!"
test[safe: 10]   // "🇺🇸"
test[11]   // "!"
test[10...]   // "🇺🇸!!! Hello Brazil 🇧🇷!!!"
test[10..<12]   // "🇺🇸!"
test[10...12]   // "🇺🇸!!"
test[...10]   // "Hello USA 🇺🇸"
test[..<10]   // "Hello USA "
test.first   // "H"
test.last    // "!"

// Subscripting the Substring
 test[...][...3]  // "Hell"

// Note that they all return a Substring of the original String.
// To create a new String from a substring
test[10...].string  // "🇺🇸!!! Hello Brazil 🇧🇷!!!"

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]

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"

Swift 5.3

I think this is very elegant. Kudos at Paul Hudson of "Hacking with Swift" for this solution:

@available (macOS 10.15, * )
extension String {
    subscript(idx: Int) -> String {
        String(self[index(startIndex, offsetBy: idx)])
    }
}

Then to get one character out of the String you simply do:

var string = "Hello, world!"

var firstChar = string[0] // No error, returns "H" as a String

NB: I just wanted to add, this will return a String as pointed out in the comments. I think it might be unexpected for Swift users, but often I need a String to use in my code straight away and not a Character type, so it does simplify my code a little bit avoiding a conversion from Character to String later.

Swift 2.0 as of Xcode 7 GM Seed

var text = "Hello, world!"

let firstChar = text[text.startIndex.advancedBy(0)] // "H"

For the nth character, replace 0 with n-1.

Edit: Swift 3.0

text[text.index(text.startIndex, offsetBy: 0)]

n.b. there are simpler ways of grabbing certain characters in the string

e.g. let firstChar = text.characters.first

If you see Cannot subscript a value of type 'String'... use this extension:

Swift 3

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 = index(startIndex, offsetBy: r.lowerBound)
        let end = index(startIndex, offsetBy: r.upperBound)
        return self[start..<end]
    }

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

Swift 2.3

extension String {
    subscript(integerIndex: Int) -> Character {
        let index = advance(startIndex, integerIndex)
        return self[index]
    }

    subscript(integerRange: Range<Int>) -> String {
        let start = advance(startIndex, integerRange.startIndex)
        let end = advance(startIndex, integerRange.endIndex)
        let range = start..<end
        return self[range]
    }
}

Source: http://oleb.net/blog/2014/07/swift-strings/

Swift 2.2 Solution:

The following extension works in Xcode 7, this is a combination of this solution and Swift 2.0 syntax conversion.

extension String {
    subscript(integerIndex: Int) -> Character {
        let index = startIndex.advancedBy(integerIndex)
        return self[index]
    }

    subscript(integerRange: Range<Int>) -> String {
        let start = startIndex.advancedBy(integerRange.startIndex)
        let end = startIndex.advancedBy(integerRange.endIndex)
        let range = start..<end
        return self[range]
    }
}

The swift string class does not provide the ability to get a character at a specific index because of its native support for UTF characters. The variable length of a UTF character in memory makes jumping directly to a character impossible. That means you have to manually loop over the string each time.

You can extend String to provide a method that will loop through the characters until your desired index

extension String {
    func characterAtIndex(index: Int) -> Character? {
        var cur = 0
        for char in self {
            if cur == index {
                return char
            }
            cur++
        }
        return nil
    }
}

myString.characterAtIndex(0)!

You can do it by convert String into Array and get it by specific index using subscript as below

var str = "Hello"
let s = Array(str)[2]
print(s)

Swift 4

String(Array(stringToIndex)[index]) 

This is probably the best way of solving this problem one-time. You probably want to cast the String as an array first, and then cast the result as a String again. Otherwise, a Character will be returned instead of a String.

Example String(Array("HelloThere")[1]) will return "e" as a String.

(Array("HelloThere")[1] will return "e" as a Character.

Swift does not allow Strings to be indexed like arrays, but this gets the job done, brute-force style.

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" 

:-)

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)]

I just had the same issue. Simply do this:

var aString: String = "test"
var aChar:unichar = (aString as NSString).characterAtIndex(0)

We have subscript which will very useful here

But String subscript will take param as String.Index so most of the people gets confuse here how to pass String.Index to get details how to form String.Index as per our requirement please look at below documentation Apple Documentation

Here i have created one extension method to get nth character in string

extension String {
    subscript(i: Int) -> String {
        return  i < count ? String(self[index(startIndex, offsetBy: i)]) : ""
    }
}

Usage

let name = "Narayana Rao Routhu"
print(name[11]) //o
print(name[1]) //a
print(name[0]) //N
print(name[30]) //""

if you pass index which is out of bounds of String count it will return empty String

By now, subscript(_:) is unavailable. As well as we can't do this

str[0] 

with string.We have to provide "String.Index" But, how can we give our own index number in this way, instead we can use,

string[str.index(str.startIndex, offsetBy: 0)]

In Swift 5 without extension to the String :

var str = "ABCDEFGH"
for char in str {
if(char == "C") { }
}

Above Swift code as same as that Java code :

int n = 8;
var str = "ABCDEFGH"
for (int i=0; i<n; i++) {
if (str.charAt(i) == 'C') { }
}

Swift 5.1.3:

Add a String extension:

extension String {

 func stringAt(_ i: Int) -> String { 
   return String(Array(self)[i]) 
 } 

 func charAt(_ i: Int) -> Character { 
  return Array(self)[i] 
 } 
}

let str = "Teja Kumar"
let str1: String = str.stringAt(2)  //"j"
let str2: Character = str.charAt(5)  //"k"

Swift 4.2 or later

Range and partial range subscripting using String's indices property

As variation of @LeoDabus nice answer, we may add an additional extension to DefaultIndices with the purpose of allowing us to fall back on the indices property of String when implementing the custom subscripts (by Int specialized ranges and partial ranges) for the latter.

extension DefaultIndices {
    subscript(at: Int) -> Elements.Index { index(startIndex, offsetBy: at) }
}

// Moving the index(_:offsetBy:) to an extension yields slightly
// briefer implementations for these String extensions.
extension String {
    subscript(range: Range<Int>) -> SubSequence {
        let start = indices[range.lowerBound]
        return self[start..<indices[start...][range.count]]
    }
    subscript(range: ClosedRange<Int>) -> SubSequence {
        let start = indices[range.lowerBound]
        return self[start...indices[start...][range.count]]
    }
    subscript(range: PartialRangeFrom<Int>) -> SubSequence {
        self[indices[range.lowerBound]...]
    }
    subscript(range: PartialRangeThrough<Int>) -> SubSequence {
        self[...indices[range.upperBound]]
    }
    subscript(range: PartialRangeUpTo<Int>) -> SubSequence {
        self[..<indices[range.upperBound]]
    }
}

let str = "foo bar baz bax"
print(str[4..<6]) // "ba"
print(str[4...6]) // "bar"
print(str[4...])  // "bar baz bax"
print(str[...6])  // "foo bar"
print(str[..<6])  // "foo ba"

Thanks @LeoDabus for the pointing me in the direction of using the indices property as an(other) alternative to String subscripting!

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.

Swift3

You can use subscript syntax to access the Character at a particular String index.

let greeting = "Guten Tag!"
let index = greeting.index(greeting.startIndex, offsetBy: 7)
greeting[index] // a

Visit https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/StringsAndCharacters.html

or we can do a String Extension in Swift 4

extension String {
    func getCharAtIndex(_ index: Int) -> Character {
        return self[self.index(self.startIndex, offsetBy: index)]
    }
}

USAGE:

let foo = "ABC123"
foo.getCharAtIndex(2) //C

Best way which worked for me is:

var firstName = "Olivia"
var lastName = "Pope"

var nameInitials.text = "\(firstName.prefix(1))" + "\    (lastName.prefix(1))"

Output:"OP"

Here's an extension you can use, working with Swift 3.1. A single index will return a Character, which seems intuitive when indexing a String, and a Range will return a String.

extension String {
    subscript (i: Int) -> Character {
        return Array(self.characters)[i]
    }
    
    subscript (r: CountableClosedRange<Int>) -> String {
        return String(Array(self.characters)[r])
    }
    
    subscript (r: CountableRange<Int>) -> String {
        return self[r.lowerBound...r.upperBound-1]
    }
}

Some examples of the extension in action:

let string = "Hello"

let c1 = string[1]  // Character "e"
let c2 = string[-1] // fatal error: Index out of range

let r1 = string[1..<4] // String "ell"
let r2 = string[1...4] // String "ello"
let r3 = string[1...5] // fatal error: Array index is out of range

n.b. You could add an additional method to the above extension to return a String with a single character if wanted:

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

Note that then you would have to explicitly specify the type you wanted when indexing the string:

let c: Character = string[3] // Character "l"
let s: String = string[0]    // String "H"

Swift 3: another solution (tested in playground)

extension String {
    func substr(_ start:Int, length:Int=0) -> String? {
        guard start > -1 else {
            return nil
        }

        let count = self.characters.count - 1

        guard start <= count else {
            return nil
        }

        let startOffset = max(0, start)
        let endOffset = length > 0 ? min(count, startOffset + length - 1) : count

        return self[self.index(self.startIndex, offsetBy: startOffset)...self.index(self.startIndex, offsetBy: endOffset)]
    }
}

Usage:

let txt = "12345"

txt.substr(-1) //nil
txt.substr(0) //"12345"
txt.substr(0, length: 0) //"12345"
txt.substr(1) //"2345"
txt.substr(2) //"345"
txt.substr(3) //"45"
txt.substr(4) //"5"
txt.substr(6) //nil
txt.substr(0, length: 1) //"1"
txt.substr(1, length: 1) //"2"
txt.substr(2, length: 1) //"3"
txt.substr(3, length: 1) //"4"
txt.substr(3, length: 2) //"45"
txt.substr(3, length: 3) //"45"
txt.substr(4, length: 1) //"5"
txt.substr(4, length: 2) //"5"
txt.substr(5, length: 1) //nil
txt.substr(5, length: -1) //nil
txt.substr(-1, length: -1) //nil

Get & Set Subscript (String & Substring) - Swift 4.2

Swift 4.2, Xcode 10

I based my answer off of @alecarlson's answer. The only big difference is you can get a Substring or a String returned (and in some cases, a single Character). You can also get and set the subscript. Lastly, mine is a bit more cumbersome and longer than @alecarlson's answer and as such, I suggest you put it in a source file.

Extension:

public extension String {
    public subscript (i: Int) -> Character {
        get {
            return self[index(startIndex, offsetBy: i)]
        }
        set (c) {
            let n = index(startIndex, offsetBy: i)
            replaceSubrange(n...n, with: "\(c)")
        }
    }
    public subscript (bounds: CountableRange<Int>) -> Substring {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[start ..< end]
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ..< end, with: s)
        }
    }
    public subscript (bounds: CountableClosedRange<Int>) -> Substring {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[start ... end]
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ... end, with: s)
        }
        
    }
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> Substring {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            return self[start ... end]
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            replaceSubrange(start ... end, with: s)
        }
    }
    public subscript (bounds: PartialRangeThrough<Int>) -> Substring {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[startIndex ... end]
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ... end, with: s)
        }
    }
    public subscript (bounds: PartialRangeUpTo<Int>) -> Substring {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[startIndex ..< end]
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ..< end, with: s)
        }
    }
    
    public subscript (i: Int) -> String {
        get {
            return "\(self[index(startIndex, offsetBy: i)])"
        }
        set (c) {
            let n = index(startIndex, offsetBy: i)
            self.replaceSubrange(n...n, with: "\(c)")
        }
    }
    public subscript (bounds: CountableRange<Int>) -> String {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[start ..< end])"
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ..< end, with: s)
        }
    }
    public subscript (bounds: CountableClosedRange<Int>) -> String {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[start ... end])"
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ... end, with: s)
        }
        
    }
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> String {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            return "\(self[start ... end])"
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            replaceSubrange(start ... end, with: s)
        }
    }
    public subscript (bounds: PartialRangeThrough<Int>) -> String {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[startIndex ... end])"
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ... end, with: s)
        }
    }
    public subscript (bounds: PartialRangeUpTo<Int>) -> String {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[startIndex ..< end])"
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ..< end, with: s)
        }
    }
    
    public subscript (i: Int) -> Substring {
        get {
            return Substring("\(self[index(startIndex, offsetBy: i)])")
        }
        set (c) {
            let n = index(startIndex, offsetBy: i)
            replaceSubrange(n...n, with: "\(c)")
        }
    }
}
public extension Substring {
    public subscript (i: Int) -> Character {
        get {
            return self[index(startIndex, offsetBy: i)]
        }
        set (c) {
            let n = index(startIndex, offsetBy: i)
            replaceSubrange(n...n, with: "\(c)")
        }
        
    }
    public subscript (bounds: CountableRange<Int>) -> Substring {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[start ..< end]
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ..< end, with: s)
        }
    }
    public subscript (bounds: CountableClosedRange<Int>) -> Substring {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[start ... end]
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ... end, with: s)
        }
    }
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> Substring {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            return self[start ... end]
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            replaceSubrange(start ... end, with: s)
        }
        
    }
    public subscript (bounds: PartialRangeThrough<Int>) -> Substring {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[startIndex ... end]
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ..< end, with: s)
        }
    }
    public subscript (bounds: PartialRangeUpTo<Int>) -> Substring {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return self[startIndex ..< end]
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ..< end, with: s)
        }
    }
    public subscript (i: Int) -> String {
        get {
            return "\(self[index(startIndex, offsetBy: i)])"
        }
        set (c) {
            let n = index(startIndex, offsetBy: i)
            replaceSubrange(n...n, with: "\(c)")
        }
    }
    public subscript (bounds: CountableRange<Int>) -> String {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[start ..< end])"
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ..< end, with: s)
        }
    }
    public subscript (bounds: CountableClosedRange<Int>) -> String {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[start ... end])"
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(start ... end, with: s)
        }
        
    }
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> String {
        get {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            return "\(self[start ... end])"
        }
        set (s) {
            let start = index(startIndex, offsetBy: bounds.lowerBound)
            let end = index(endIndex, offsetBy: -1)
            replaceSubrange(start ... end, with: s)
        }
    }
    public subscript (bounds: PartialRangeThrough<Int>) -> String {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[startIndex ... end])"
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ... end, with: s)
        }
    }
    public subscript (bounds: PartialRangeUpTo<Int>) -> String {
        get {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            return "\(self[startIndex ..< end])"
        }
        set (s) {
            let end = index(startIndex, offsetBy: bounds.upperBound)
            replaceSubrange(startIndex ..< end, with: s)
        }
    }
    
    public subscript (i: Int) -> Substring {
        get {
            return Substring("\(self[index(startIndex, offsetBy: i)])")
        }
        set (c) {
            let n = index(startIndex, offsetBy: i)
            replaceSubrange(n...n, with: "\(c)")
        }
    }
}

Swift 4.2

This answer is ideal because it extends String and all of its Subsequences (Substring) in one extension

public extension StringProtocol {
    
    public subscript (i: Int) -> Element {
        return self[index(startIndex, offsetBy: i)]
    }

    public subscript (bounds: CountableClosedRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[start...end]
    }
    
    public subscript (bounds: CountableRange<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[start..<end]
    }
    
    public subscript (bounds: PartialRangeUpTo<Int>) -> SubSequence {
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[startIndex..<end]
    }
    
    public subscript (bounds: PartialRangeThrough<Int>) -> SubSequence {
        let end = index(startIndex, offsetBy: bounds.upperBound)
        return self[startIndex...end]
    }
    
    public subscript (bounds: CountablePartialRangeFrom<Int>) -> SubSequence {
        let start = index(startIndex, offsetBy: bounds.lowerBound)
        return self[start..<endIndex]
    }
}

Usage

var str = "Hello, playground"

print(str[5...][...5][0])
// Prints ","

Update for swift 2.0 subString

public extension String {
    public subscript (i: Int) -> String {
        return self.substringWithRange(self.startIndex..<self.startIndex.advancedBy(i + 1))
    }

    public subscript (r: Range<Int>) -> String {
        get {
            return self.substringWithRange(self.startIndex.advancedBy(r.startIndex)..<self.startIndex.advancedBy(r.endIndex))
        }
    }

}

I think that a fast answer for get the first character could be:

let firstCharacter = aString[aString.startIndex]

It's so much elegant and performance than:

let firstCharacter = Array(aString.characters).first

But.. if you want manipulate and do more operations with strings you could think create an extension..here is one extension with this approach, it's quite similar to that already posted here:

extension String {
var length : Int {
    return self.characters.count
}

subscript(integerIndex: Int) -> Character {
    let index = startIndex.advancedBy(integerIndex)
    return self[index]
}

subscript(integerRange: Range<Int>) -> String {
    let start = startIndex.advancedBy(integerRange.startIndex)
    let end = startIndex.advancedBy(integerRange.endIndex)
    let range = start..<end
    return self[range]
}

}

BUT IT'S A TERRIBLE IDEA!!

The extension below is horribly inefficient. Every time a string is accessed with an integer, an O(n) function to advance its starting index is run. Running a linear loop inside another linear loop means this for loop is accidentally O(n2) — as the length of the string increases, the time this loop takes increases quadratically.

Instead of doing that you could use the characters's string collection.