Published Feb 19, 2022Updated Oct 11, 2022
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Optionals are types that deal with scenarios involving the potential presence or absence of a value (nil).


In Swift, a non-optional variable should not be set to nil. Rather, the optional type must be defined using the added question mark ?.

There are two ways to use an optional. Below, a variable named myVariable is set to an optional Type, Type?. Type referred to a Swift data type. It can either be equal to some value with the specified data type or nil.

var myVariable: Type?
var myVariable: Optional<Type>

Note: If myVariable is not assigned a value of the specified data type, it will default to nil.

Unwrapping Optionals

Optional types should be unwrapped only if it’s certain that a value is present. An error will result if an optional without a value is unwrapped. Below are some ways this can be done.


The nil-coalescing operator, ??, assigns a value by checking from left to right. If the preceding optional values are nil, the default value on the right is assigned.

In the example below,maybeInt is declared as an optional integer type and set to nil, therefore myInt will default to 27:

var maybeInt: Int? = nil
var myInt = maybeInt ?? 27
// Output: 27

Note: In the following code blocks, a mutable var variable can be used in place of any constant let variable.

if let

if let is an optional binding control structure.

var maybeString: String? = "maybeString"
if let myString = maybeString {
print("\(maybeString) and myString are equal!")
} else {
print("myString could not be assigned to nil.")
// Output: Optional("maybeString") and myString are equal!

myString is declared and conditionally bound to the maybeString optional. The value of myString can be accessed from inside the if statement, but not in the outer scope. An alternative to if let that allows upwrapped values to be accessed in the outer scope, is the guard let statement.

guard let

guard let is another optional binding control structure that can be used within a function.

var maybeInt: Int? = 2022
func findInt() {
guard let myInt = maybeInt else {
// Output: 2022

If the optional maybeInt holds a value, it will be assigned to myInt. Otherwise, the else block will run and hit the return statement.

Note: The following way(s) to unwrap optionals is not recommended because it does not handle nil cases.

Forced Upwrapping

Placing an ! after an optional value forces it to unwrap its value:

var maybeInt: Int? = 27
var myInt: Int = maybeInt!
// Output: 27

Alternatively, this can be done implicitly:

var myInt = maybeInt!
// Output: 27

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