Swift Language Optionals


“ An optional value either contains a value or contains nil to indicate that a value is missing”

Excerpt From: Apple Inc. “The Swift Programming Language (Swift 3.1 Edition).” iBooks. https://itun.es/us/k5SW7.l

Basic optional use cases include: for a constant (let), use of an optional within a loop (if-let), safely unwrapping an optional value within a method (guard-let), and as part of switch loops (case-let), defaulting to a value if nil, using the coalesce operator (??)


  • var optionalName: optionalType? // declare an optional type, defaults to nil
  • var optionalName: optionalType? = value // declare an optional with a value
  • var optionalName: optionalType! // declare an implicitly unwrapped optional
  • optional! // force unwrap an optional


For more information about optionals, see The Swift Programming Language.

Types of Optionals

Optionals are a generic enum type that acts as a wrapper. This wrapper allows a variable to have one of two states: the value of the user-defined type or nil, which represents the absence of a value.

This ability is particularly important in Swift because one of the stated design objectives of the language is to work well with Apple's frameworks. Many (most) of Apple's frameworks utilize nil due to its ease of use and significance to programming patterns and API design within Objective-C.

In Swift, for a variable to have a nil value, it must be an optional. Optionals can be created by appending either a ! or a ? to the variable type. For example, to make an Int optional, you could use

var numberOne: Int! = nil
var numberTwo: Int? = nil

? optionals must be explicitly unwrapped, and should be used if you aren't certain whether or not the variable will have a value when you access it. For example, when turning a string into an Int, the result is an optional Int?, because nil will be returned if the string is not a valid number

let str1 = "42"
let num1: Int? = Int(str1) // 42

let str2 = "Hello, World!"
let num2: Int? = Int(str2) // nil

! optionals are automatically unwrapped, and should only be used when you are certain that the variable will have a value when you access it. For example, a global UIButton! variable that is initialized in viewDidLoad()

//myButton will not be accessed until viewDidLoad is called,
//so a ! optional can be used here
var myButton: UIButton!

override func viewDidLoad(){
    self.myButton = UIButton(frame: self.view.frame)
    self.myButton.backgroundColor = UIColor.redColor()

Unwrapping an Optional

In order to access the value of an Optional, it needs to be unwrapped.

You can conditionally unwrap an Optional using optional binding and force unwrap an Optional using the ! operator.

Conditionally unwrapping effectively asks "Does this variable have a value?" while force unwrapping says "This variable has a value!".

If you force unwrap a variable that is nil, your program will throw an unexpectedly found nil while unwrapping an optional exception and crash, so you need to consider carefully if using ! is appropriate.

var text: String? = nil
var unwrapped: String = text! //crashes with "unexpectedly found nil while unwrapping an Optional value"

For safe unwrapping, you can use an if-let statement, which will not throw an exception or crash if the wrapped value is nil:

var number: Int?
if let unwrappedNumber = number {       // Has `number` been assigned a value?
    print("number: \(unwrappedNumber)") // Will not enter this line
} else {
    print("number was not assigned a value")

Or, a guard statement:

var number: Int?
guard let unwrappedNumber = number else {
print("number: \(unwrappedNumber)")

Note that the scope of the unwrappedNumber variable is inside the if-let statement and outside of the guard block.

You can chain unwrapping of many optionals, this is mainly useful in cases that your code requires more then variable to run correctly:

var firstName:String?
var lastName:String?

if let fn = firstName, let ln = lastName {
    print("\(fn) + \(ln)")//pay attention that the condition will be true only if both optionals are not nil.

Note that all the variables have to be unwrapped in order to pass successfully the test, otherwise you would have no way to determine which variables were unwrapped and which weren't.

You can chain conditional statements using your optionals immediately after it is unwrapped. This means no nested if - else statements!

var firstName:String? = "Bob"
var myBool:Bool? = false

if let fn = firstName, fn == "Bob", let bool = myBool, !bool {
    print("firstName is bob and myBool was false!")

Nil Coalescing Operator

You can use the nil coalescing operator to unwrap a value if it is non-nil, otherwise provide a different value:

func fallbackIfNil(str: String?) -> String {
    return str ?? "Fallback String"
print(fallbackIfNil("Hi")) // Prints "Hi"
print(fallbackIfNil(nil)) // Prints "Fallback String"

This operator is able to short-circuit, meaning that if the left operand is non-nil, the right operand will not be evaluated:

func someExpensiveComputation() -> String { ... }

var foo : String? = "a string"
let str = foo ?? someExpensiveComputation()

In this example, as foo is non-nil, someExpensiveComputation() will not be called.

You can also chain multiple nil coalescing statements together:

var foo : String?
var bar : String?

let baz = foo ?? bar ?? "fallback string"

In this example baz will be assigned the unwrapped value of foo if it is non-nil, otherwise it will be assigned the unwrapped value of bar if it is non-nil, otherwise it will be assigned the fallback value.

Optional Chaining

You can use Optional Chaining in order to call a method, access a property or subscript an optional. This is done by placing a ? between the given optional variable and the given member (method, property or subscript).

struct Foo {
    func doSomething() {
        print("Hello World!")

var foo : Foo? = Foo()

foo?.doSomething() // prints "Hello World!" as foo is non-nil

If foo contains a value, doSomething() will be called on it. If foo is nil, then nothing bad will happen – the code will simply fail silently and continue executing.

var foo : Foo? = nil

foo?.doSomething() // will not be called as foo is nil

(This is similar behaviour to sending messages to nil in Objective-C)

The reason that Optional Chaining is named as such is because ‘optionality’ will be propagated through the members you call/access. What this means is that the return values of any members used with optional chaining will be optional, regardless of whether they are typed as optional or not.

struct Foo {
    var bar : Int
    func doSomething() { ... }

let foo : Foo? = Foo(bar: 5)
print(foo?.bar) // Optional(5)

Here foo?.bar is returning an Int? even though bar is non-optional, as foo itself is optional.

As optionality is propagated, methods returning Void will return Void? when called with optional chaining. This can be useful in order to determine whether the method was called or not (and therefore if the optional has a value).

let foo : Foo? = Foo()

if foo?.doSomething() != nil {
    print("foo is non-nil, and doSomething() was called")
} else {
    print("foo is nil, therefore doSomething() wasn't called")

Here we’re comparing the Void? return value with nil in order to determine whether the method was called (and therefore whether foo is non-nil).

Overview - Why Optionals?

Often when programming it is necessary to make some distinction between a variable that has a value and one that does not. For reference types, such as C Pointers, a special value such as null can be used to indicate that the variable has no value. For intrinsic types, such as an integer, it is more difficult. A nominated value, such as -1 can be used, but this relies on interpretation of the value. It also eliminates that "special" value from normal use.

To address this, Swift allows any variable to be declared as an optional. This is indicated by the use of a ? or ! after the type (See Types of optionals)

For example,

var possiblyInt: Int?

declares a variable that may or may not contain an integer value.

The special value nil indicates that no value is currently assigned to this variable.

possiblyInt = 5      // PossiblyInt is now 5
possiblyInt = nil    // PossiblyInt is now unassigned

nil can also be used to test for an assigned value:

if possiblyInt != nil {
    print("possiblyInt has the value \(possiblyInt!)")

Note the use of ! in the print statement to unwrap the optional value.

As an example of a common use of optionals, consider a function that returns an integer from a string containing digits; It is possible that the string may contain non-digit characters, or may even be empty.

How can a function that returns a simple Int indicate failure? It cannot do so by returning any specific value as this would preclude that value from being parsed from the string.

var someInt
someInt = parseInt("not an integer") // How would this function indicate failure?

In Swift, however, that function can simply return an optional Int. Then failure is indicated by return value of nil.

var someInt?
someInt = parseInt("not an integer")  // This function returns nil if parsing fails
if someInt == nil {
    print("That isn't a valid integer")