Optionals in

About Optionals

Optionals

Swift uses optionals to allow programmers to represent the possible absence of a value. Optional is a type that can either hold a value or be nil, which represents the absence of a value. Using an optional requires a program to check if a value does exist before using it after unwrapping it.

Any type can be made into an optional by appending a ? onto the end of the type name. So an optional integer would have type Int? and an optional string would have type String?. Defining constants or variables of optional type and assigning them values is done the same as for values of non-optional types.

let x: Int? = 42
var y: String? = "Hello"
y = "Goodbye"

You can assign the absence of a value to a variable of optional type by assigning it the special value nil. nil can be used with all optional types, but nils assigned to two different optional types do not have the same type, and cannot be interchanged or even compared.

let intOpt: Int? = nil
let stringOpt: String? = nil

intOpt = stringOpt
// Compiler error: Cannot assign value of type 'String?' to type 'Int?'

intOpt == stringOpt
// Compiler error: Binary operator '==' cannot be applied to operands of type 'Int?' and 'String?'

Also note that even though nil can be assigned to any optional type, it cannot be assigned to a non-optional type (even if it doesn't actually hold nil). And methods that is expecting a non-optional type cannot be passed an optional type without unwrapping it first.

var x: Int = 42
var y: Int? = 42

y = x
// Works fine
x = y
// error: value of optional type 'Int?' must be unwrapped to a value of type 'Int'
x + y
// error: value of optional type 'Int?' must be unwrapped to a value of type 'Int'

Using optionals

Because optional types are not the same types as their base types, the two types cannot be used in the same ways. For example: Int("123") + 1 results in a compiler error "Value of optional type 'Int?' must be unwrapped to a value of type 'Int'". This is because the Int("123") returns an optional Int? type, not an Int type, since if the string cannot be converted to an integer, the result will be nil. In order to access the Int from the conversion, one must "unwrap" it first.

This is most commonly done in Swift using the if-let and guard-let constructs for optional binding which check for nil and take one code path with the unwrapped value bound to a supplied name if a value exists and taking a different code path if nil was found.

if let safeNum = Int("123") {
	let sum = safeNum + 1
	…
} else {
  // code for the case where nil was found -- may be left out
}

It is worth noting that the safeNum variable has the type Int and not Int?. In the example below, num is of type Int? and safeNum is of type Int.

let num = Int("123")
if let safeNum = num {
  // num is of type Int
}

This optional binding is important because it unwraps (or "removes") the optional type from the value, allowing it to be used as a non-optional value. If you would just do a conditional check on if the value is nil, the value would still be of optional type:

let num = Int("123")
if num != nil {
  // num is of type Int?
}

The guard-let option may also be used in the cases where early return is desired:

guard let num = Int("123") else { return nil }
let sum = num + 1
…

Comparing optionals

Note that even if the base type of a pair of optionals can be compared using the standard comparison operators, the optionals themselves cannot be compared. They can only be checked for equality. Two optionals are equal if they are both nil or if the values they wrap are equal within their base types.

However, code can be written to perform a custom comparison of two optional values. Below is an example of a switch statement that will return true only if both optional values are non-nil and the first value is less than the second. To do this it uses the optional pattern varName? which only matches non-nil optionals, binding the value inside the optional to the name varName.

switch (optionalA, optionalB) {
case let (valA?, valB?): return valA < valB
default: return false
}

Nil coalescing

Another option for unwrapping exists where it is possible to use a fallback value if nil is present. This can be done by using the nil coalescing operator, ??. Assuming x is an Int?, if one writes let y = x ?? 0, then Swift will check if x is nil. If it is not, then it will unwrap x and assign the unwrapped value to y, and if x is nil, then it will assign 0 to y.

Since x ?? y is simply shorthand for x != nil ? x! : y, if x is not nil, then the expression y is not evaluated at all.

Finally, it should be noted that the nil coalescing operator is right associative, which can lead to surprising results for the unaware.

let k = 42 ?? 0 + 1    // returns 42
let j = nil ?? 0 + 1   // returns 1

You can read further about the nil coalescing operator in A Tour of Swift: Nil-Coalescing Operator.