Shorthand for Wrapping a Swift Variable in an Optional

Swift: shortcut unwrapping of array of optionals

This solution will get you a new array with all values unwrapped and all nil's filtered away.

Swift 4.1:

let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.compactMap { $0 }

Swift 2.0:

let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.flatMap { $0 }

Swift 1.0:

let arrayOfOptionals: [String?] = ["Seems", "like", "an", nil, "of", "optionals"]
let arrayWithNoOptionals = arrayOfOptionals.filter { $0 != nil }.map { $0! }

What is an optional value in Swift?

An optional in Swift is a type that can hold either a value or no value. Optionals are written by appending a ? to any type:

var name: String? = "Bertie"

Optionals (along with Generics) are one of the most difficult Swift concepts to understand. Because of how they are written and used, it's easy to get a wrong idea of what they are. Compare the optional above to creating a normal String:

var name: String = "Bertie" // No "?" after String

From the syntax it looks like an optional String is very similar to an ordinary String. It's not. An optional String is not a String with some "optional" setting turned on. It's not a special variety of String. A String and an optional String are completely different types.

Here's the most important thing to know: An optional is a kind of container. An optional String is a container which might contain a String. An optional Int is a container which might contain an Int. Think of an optional as a kind of parcel. Before you open it (or "unwrap" in the language of optionals) you won't know if it contains something or nothing.

You can see how optionals are implemented in the Swift Standard Library by typing "Optional" into any Swift file and ⌘-clicking on it. Here's the important part of the definition:

enum Optional<Wrapped> {
    case none
    case some(Wrapped)
}

Optional is just an enum which can be one of two cases: .none or .some. If it's .some, there's an associated value which, in the example above, would be the String "Hello". An optional uses Generics to give a type to the associated value. The type of an optional String isn't String, it's Optional, or more precisely Optional<String>.

Everything Swift does with optionals is magic to make reading and writing code more fluent. Unfortunately this obscures the way it actually works. I'll go through some of the tricks later.

Note: I'll be talking about optional variables a lot, but it's fine to create optional constants too. I mark all variables with their type to make it easier to understand type types being created, but you don't have to in your own code.

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How to create optionals

To create an optional, append a ? after the type you wish to wrap. Any type can be optional, even your own custom types. You can't have a space between the type and the ?.

var name: String? = "Bob" // Create an optional String that contains "Bob"
var peter: Person? = Person() // An optional "Person" (custom type)

// A class with a String and an optional String property
class Car {
var modelName: String // must exist
var internalName: String? // may or may not exist
}

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Using optionals

You can compare an optional to nil to see if it has a value:

var name: String? = "Bob"
name = nil // Set name to nil, the absence of a value
if name != nil {
    print("There is a name")
}
if name == nil { // Could also use an "else"
    print("Name has no value")
}

This is a little confusing. It implies that an optional is either one thing or another. It's either nil or it's "Bob". This is not true, the optional doesn't transform into something else. Comparing it to nil is a trick to make easier-to-read code. If an optional equals nil, this just means that the enum is currently set to .none.

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Only optionals can be nil

If you try to set a non-optional variable to nil, you'll get an error.

var red: String = "Red"
red = nil // error: nil cannot be assigned to type 'String'

Another way of looking at optionals is as a complement to normal Swift variables. They are a counterpart to a variable which is guaranteed to have a value. Swift is a careful language that hates ambiguity. Most variables are define as non-optionals, but sometimes this isn't possible. For example, imagine a view controller which loads an image either from a cache or from the network. It may or may not have that image at the time the view controller is created. There's no way to guarantee the value for the image variable. In this case you would have to make it optional. It starts as nil and when the image is retrieved, the optional gets a value.

Using an optional reveals the programmers intent. Compared to Objective-C, where any object could be nil, Swift needs you to be clear about when a value can be missing and when it's guaranteed to exist.

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To use an optional, you "unwrap" it

An optional String cannot be used in place of an actual String. To use the wrapped value inside an optional, you have to unwrap it. The simplest way to unwrap an optional is to add a ! after the optional name. This is called "force unwrapping". It returns the value inside the optional (as the original type) but if the optional is nil, it causes a runtime crash. Before unwrapping you should be sure there's a value.

var name: String? = "Bob"
let unwrappedName: String = name!
print("Unwrapped name: \(unwrappedName)")

name = nil
let nilName: String = name! // Runtime crash. Unexpected nil.

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Checking and using an optional

Because you should always check for nil before unwrapping and using an optional, this is a common pattern:

var mealPreference: String? = "Vegetarian"
if mealPreference != nil {
    let unwrappedMealPreference: String = mealPreference!
    print("Meal: \(unwrappedMealPreference)") // or do something useful
}

In this pattern you check that a value is present, then when you are sure it is, you force unwrap it into a temporary constant to use. Because this is such a common thing to do, Swift offers a shortcut using "if let". This is called "optional binding".

var mealPreference: String? = "Vegetarian"
if let unwrappedMealPreference: String = mealPreference {
    print("Meal: \(unwrappedMealPreference)") 
}

This creates a temporary constant (or variable if you replace let with var) whose scope is only within the if's braces. Because having to use a name like "unwrappedMealPreference" or "realMealPreference" is a burden, Swift allows you to reuse the original variable name, creating a temporary one within the bracket scope

var mealPreference: String? = "Vegetarian"
if let mealPreference: String = mealPreference {
    print("Meal: \(mealPreference)") // separate from the other mealPreference
}

Here's some code to demonstrate that a different variable is used:

var mealPreference: String? = "Vegetarian"
if var mealPreference: String = mealPreference {
    print("Meal: \(mealPreference)") // mealPreference is a String, not a String?
    mealPreference = "Beef" // No effect on original
}
// This is the original mealPreference
print("Meal: \(mealPreference)") // Prints "Meal: Optional("Vegetarian")"

Optional binding works by checking to see if the optional equals nil. If it doesn't, it unwraps the optional into the provided constant and executes the block. In Xcode 8.3 and later (Swift 3.1), trying to print an optional like this will cause a useless warning. Use the optional's debugDescription to silence it:

print("\(mealPreference.debugDescription)")

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What are optionals for?

Optionals have two use cases:

1. Things that can fail (I was expecting something but I got nothing)
2. Things that are nothing now but might be something later (and vice-versa)

Some concrete examples:

• A property which can be there or not there, like middleName or spouse in a Person class
• A method which can return a value or nothing, like searching for a match in an array
• A method which can return either a result or get an error and return nothing, like trying to read a file's contents (which normally returns the file's data) but the file doesn't exist
• Delegate properties, which don't always have to be set and are generally set after initialization
• For weak properties in classes. The thing they point to can be set to nil at any time
• A large resource that might have to be released to reclaim memory
• When you need a way to know when a value has been set (data not yet loaded > the data) instead of using a separate dataLoaded Boolean

Optionals don't exist in Objective-C but there is an equivalent concept, returning nil. Methods that can return an object can return nil instead. This is taken to mean "the absence of a valid object" and is often used to say that something went wrong. It only works with Objective-C objects, not with primitives or basic C-types (enums, structs). Objective-C often had specialized types to represent the absence of these values (NSNotFound which is really NSIntegerMax, kCLLocationCoordinate2DInvalid to represent an invalid coordinate, -1 or some negative value are also used). The coder has to know about these special values so they must be documented and learned for each case. If a method can't take nil as a parameter, this has to be documented. In Objective-C, nil was a pointer just as all objects were defined as pointers, but nil pointed to a specific (zero) address. In Swift, nil is a literal which means the absence of a certain type.

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Comparing to nil

You used to be able to use any optional as a Boolean:

let leatherTrim: CarExtras? = nil
if leatherTrim {
    price = price + 1000
}

In more recent versions of Swift you have to use leatherTrim != nil. Why is this? The problem is that a Boolean can be wrapped in an optional. If you have Boolean like this:

var ambiguous: Boolean? = false

it has two kinds of "false", one where there is no value and one where it has a value but the value is false. Swift hates ambiguity so now you must always check an optional against nil.

You might wonder what the point of an optional Boolean is? As with other optionals the .none state could indicate that the value is as-yet unknown. There might be something on the other end of a network call which takes some time to poll. Optional Booleans are also called "Three-Value Booleans"

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Swift tricks

Swift uses some tricks to allow optionals to work. Consider these three lines of ordinary looking optional code;

var religiousAffiliation: String? = "Rastafarian"
religiousAffiliation = nil
if religiousAffiliation != nil { ... }

None of these lines should compile.

• The first line sets an optional String using a String literal, two different types. Even if this was a String the types are different
• The second line sets an optional String to nil, two different types
• The third line compares an optional string to nil, two different types

I'll go through some of the implementation details of optionals that allow these lines to work.

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Creating an optional

Using ? to create an optional is syntactic sugar, enabled by the Swift compiler. If you want to do it the long way, you can create an optional like this:

var name: Optional<String> = Optional("Bob")

This calls Optional's first initializer, public init(_ some: Wrapped), which infers the optional's associated type from the type used within the parentheses.

The even longer way of creating and setting an optional:

var serialNumber:String? = Optional.none
serialNumber = Optional.some("1234")
print("\(serialNumber.debugDescription)")

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Setting an optional to nil

You can create an optional with no initial value, or create one with the initial value of nil (both have the same outcome).

var name: String?
var name: String? = nil

Allowing optionals to equal nil is enabled by the protocol ExpressibleByNilLiteral (previously named NilLiteralConvertible). The optional is created with Optional's second initializer, public init(nilLiteral: ()). The docs say that you shouldn't use ExpressibleByNilLiteral for anything except optionals, since that would change the meaning of nil in your code, but it's possible to do it:

class Clint: ExpressibleByNilLiteral {
    var name: String?
    required init(nilLiteral: ()) {
        name = "The Man with No Name"
    }
}

let clint: Clint = nil // Would normally give an error
print("\(clint.name)")

The same protocol allows you to set an already-created optional to nil. Although it's not recommended, you can use the nil literal initializer directly:

var name: Optional<String> = Optional(nilLiteral: ())

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Comparing an optional to nil

Optionals define two special "==" and "!=" operators, which you can see in the Optional definition. The first == allows you to check if any optional is equal to nil. Two different optionals which are set to .none will always be equal if the associated types are the same. When you compare to nil, behind the scenes Swift creates an optional of the same associated type, set to .none then uses that for the comparison.

// How Swift actually compares to nil
var tuxedoRequired: String? = nil
let temp: Optional<String> = Optional.none
if tuxedoRequired == temp { // equivalent to if tuxedoRequired == nil
    print("tuxedoRequired is nil")
}

The second == operator allows you to compare two optionals. Both have to be the same type and that type needs to conform to Equatable (the protocol which allows comparing things with the regular "==" operator). Swift (presumably) unwraps the two values and compares them directly. It also handles the case where one or both of the optionals are .none. Note the distinction between comparing to the nil literal.

Furthermore, it allows you to compare any Equatable type to an optional wrapping that type:

let numberToFind: Int = 23
let numberFromString: Int? = Int("23") // Optional(23)
if numberToFind == numberFromString {
    print("It's a match!") // Prints "It's a match!"
}

Behind the scenes, Swift wraps the non-optional as an optional before the comparison. It works with literals too (if 23 == numberFromString {)

I said there are two == operators, but there's actually a third which allow you to put nil on the left-hand side of the comparison

if nil == name { ... }

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Naming Optionals

There is no Swift convention for naming optional types differently from non-optional types. People avoid adding something to the name to show that it's an optional (like "optionalMiddleName", or "possibleNumberAsString") and let the declaration show that it's an optional type. This gets difficult when you want to name something to hold the value from an optional. The name "middleName" implies that it's a String type, so when you extract the String value from it, you can often end up with names like "actualMiddleName" or "unwrappedMiddleName" or "realMiddleName". Use optional binding and reuse the variable name to get around this.

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The official definition

From "The Basics" in the Swift Programming Language:

Swift also introduces optional types, which handle the absence of a value. Optionals say either “there is a value, and it equals x” or “there isn’t a value at all”. Optionals are similar to using nil with pointers in Objective-C, but they work for any type, not just classes. Optionals are safer and more expressive than nil pointers in Objective-C and are at the heart of many of Swift’s most powerful features.

Optionals are an example of the fact that Swift is a type safe language. Swift helps you to be clear about the types of values your code can work with. If part of your code expects a String, type safety prevents you from passing it an Int by mistake. This enables you to catch and fix errors as early as possible in the development process.

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To finish, here's a poem from 1899 about optionals:

Yesterday upon the stair

I met a man who wasn’t there

He wasn’t there again today

I wish, I wish he’d go away

Antigonish

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More resources:

• The Swift Programming Guide
• Optionals in Swift (Medium)
• WWDC Session 402 "Introduction to Swift" (starts around 14:15)
• More optional tips and tricks

What does Fatal error: Unexpectedly found nil while unwrapping an Optional value mean?

Background: What’s an Optional?

In Swift, Optional<Wrapped> is an option type: it can contain any value from the original ("Wrapped") type, or no value at all (the special value nil). An optional value must be unwrapped before it can be used.

Optional is a generic type, which means that Optional<Int> and Optional<String> are distinct types — the type inside <> is called the Wrapped type. Under the hood, an Optional is an enum with two cases: .some(Wrapped) and .none, where .none is equivalent to nil.

Optionals can be declared using the named type Optional<T>, or (most commonly) as a shorthand with a ? suffix.

var anInt: Int = 42
var anOptionalInt: Int? = 42
var anotherOptionalInt: Int?  // `nil` is the default when no value is provided
var aVerboseOptionalInt: Optional<Int>  // equivalent to `Int?`

anOptionalInt = nil // now this variable contains nil instead of an integer

Optionals are a simple yet powerful tool to express your assumptions while writing code. The compiler can use this information to prevent you from making mistakes. From The Swift Programming Language:

Swift is a type-safe language, which means the language helps you to be clear about the types of values your code can work with. If part of your code requires a String, type safety prevents you from passing it an Int by mistake. Likewise, type safety prevents you from accidentally passing an optional String to a piece of code that requires a non-optional String. Type safety helps you catch and fix errors as early as possible in the development process.

Some other programming languages also have generic option types: for example, Maybe in Haskell, option in Rust, and optional in C++17.

In programming languages without option types, a particular "sentinel" value is often used to indicate the absence of a valid value. In Objective-C, for example, nil (the null pointer) represents the lack of an object. For primitive types such as int, a null pointer can't be used, so you would need either a separate variable (such as value: Int and isValid: Bool) or a designated sentinel value (such as -1 or INT_MIN). These approaches are error-prone because it's easy to forget to check isValid or to check for the sentinel value. Also, if a particular value is chosen as the sentinel, that means it can no longer be treated as a valid value.

Option types such as Swift's Optional solve these problems by introducing a special, separate nil value (so you don't have to designate a sentinel value), and by leveraging the strong type system so the compiler can help you remember to check for nil when necessary.

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Why did I get “Fatal error: Unexpectedly found nil while unwrapping an Optional value”?

In order to access an optional’s value (if it has one at all), you need to unwrap it. An optional value can be unwrapped safely or forcibly. If you force-unwrap an optional, and it didn't have a value, your program will crash with the above message.

Xcode will show you the crash by highlighting a line of code. The problem occurs on this line.

This crash can occur with two different kinds of force-unwrap:

1. Explicit Force Unwrapping

This is done with the ! operator on an optional. For example:

let anOptionalString: String?
print(anOptionalString!) // <- CRASH

Fatal error: Unexpectedly found nil while unwrapping an Optional value

As anOptionalString is nil here, you will get a crash on the line where you force unwrap it.

2. Implicitly Unwrapped Optionals

These are defined with a !, rather than a ? after the type.

var optionalDouble: Double!   // this value is implicitly unwrapped wherever it's used

These optionals are assumed to contain a value. Therefore whenever you access an implicitly unwrapped optional, it will automatically be force unwrapped for you. If it doesn’t contain a value, it will crash.

print(optionalDouble) // <- CRASH

Fatal error: Unexpectedly found nil while implicitly unwrapping an Optional value

In order to work out which variable caused the crash, you can hold ⌥ while clicking to show the definition, where you might find the optional type.

IBOutlets, in particular, are usually implicitly unwrapped optionals. This is because your xib or storyboard will link up the outlets at runtime, after initialization. You should therefore ensure that you’re not accessing outlets before they're loaded in. You also should check that the connections are correct in your storyboard/xib file, otherwise the values will be nil at runtime, and therefore crash when they are implicitly unwrapped. When fixing connections, try deleting the lines of code that define your outlets, then reconnect them.

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When should I ever force unwrap an Optional?

Explicit Force Unwrapping

As a general rule, you should never explicitly force unwrap an optional with the ! operator. There may be cases where using ! is acceptable – but you should only ever be using it if you are 100% sure that the optional contains a value.

While there may be an occasion where you can use force unwrapping, as you know for a fact that an optional contains a value – there is not a single place where you cannot safely unwrap that optional instead.

Implicitly Unwrapped Optionals

These variables are designed so that you can defer their assignment until later in your code. It is your responsibility to ensure they have a value before you access them. However, because they involve force unwrapping, they are still inherently unsafe – as they assume your value is non-nil, even though assigning nil is valid.

You should only be using implicitly unwrapped optionals as a last resort. If you can use a lazy variable, or provide a default value for a variable – you should do so instead of using an implicitly unwrapped optional.

However, there are a few scenarios where implicitly unwrapped optionals are beneficial, and you are still able to use various ways of safely unwrapping them as listed below – but you should always use them with due caution.

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How can I safely deal with Optionals?

The simplest way to check whether an optional contains a value, is to compare it to nil.

if anOptionalInt != nil {
    print("Contains a value!")
} else {
    print("Doesn’t contain a value.")
}

However, 99.9% of the time when working with optionals, you’ll actually want to access the value it contains, if it contains one at all. To do this, you can use Optional Binding.

Optional Binding

Optional Binding allows you to check if an optional contains a value – and allows you to assign the unwrapped value to a new variable or constant. It uses the syntax if let x = anOptional {...} or if var x = anOptional {...}, depending if you need to modify the value of the new variable after binding it.

For example:

if let number = anOptionalInt {
    print("Contains a value! It is \(number)!")
} else {
    print("Doesn’t contain a number")
}

What this does is first check that the optional contains a value. If it does, then the ‘unwrapped’ value is assigned to a new variable (number) – which you can then freely use as if it were non-optional. If the optional doesn’t contain a value, then the else clause will be invoked, as you would expect.

What’s neat about optional binding, is you can unwrap multiple optionals at the same time. You can just separate the statements with a comma. The statement will succeed if all the optionals were unwrapped.

var anOptionalInt : Int?
var anOptionalString : String?

if let number = anOptionalInt, let text = anOptionalString {
    print("anOptionalInt contains a value: \(number). And so does anOptionalString, it’s: \(text)")
} else {
    print("One or more of the optionals don’t contain a value")
}

Another neat trick is that you can also use commas to check for a certain condition on the value, after unwrapping it.

if let number = anOptionalInt, number > 0 {
    print("anOptionalInt contains a value: \(number), and it’s greater than zero!")
}

The only catch with using optional binding within an if statement, is that you can only access the unwrapped value from within the scope of the statement. If you need access to the value from outside of the scope of the statement, you can use a guard statement.

A guard statement allows you to define a condition for success – and the current scope will only continue executing if that condition is met. They are defined with the syntax guard condition else {...}.

So, to use them with an optional binding, you can do this:

guard let number = anOptionalInt else {
    return
}

^{(Note that within the guard body, you must use one of the control transfer statements in order to exit the scope of the currently executing code).}

If anOptionalInt contains a value, it will be unwrapped and assigned to the new number constant. The code after the guard will then continue executing. If it doesn’t contain a value – the guard will execute the code within the brackets, which will lead to transfer of control, so that the code immediately after will not be executed.

The real neat thing about guard statements is the unwrapped value is now available to use in code that follows the statement (as we know that future code can only execute if the optional has a value). This is a great for eliminating ‘pyramids of doom’ created by nesting multiple if statements.

For example:

guard let number = anOptionalInt else {
    return
}

print("anOptionalInt contains a value, and it’s: \(number)!")

Guards also support the same neat tricks that the if statement supported, such as unwrapping multiple optionals at the same time and using the where clause.

Whether you use an if or guard statement completely depends on whether any future code requires the optional to contain a value.

Nil Coalescing Operator

The Nil Coalescing Operator is a nifty shorthand version of the ternary conditional operator, primarily designed to convert optionals to non-optionals. It has the syntax a ?? b, where a is an optional type and b is the same type as a (although usually non-optional).

It essentially lets you say “If a contains a value, unwrap it. If it doesn’t then return b instead”. For example, you could use it like this:

let number = anOptionalInt ?? 0

This will define a number constant of Int type, that will either contain the value of anOptionalInt, if it contains a value, or 0 otherwise.

It’s just shorthand for:

let number = anOptionalInt != nil ? anOptionalInt! : 0

Optional Chaining

You can use Optional Chaining in order to call a method or access a property on an optional. This is simply done by suffixing the variable name with a ? when using it.

For example, say we have a variable foo, of type an optional Foo instance.

var foo : Foo?

If we wanted to call a method on foo that doesn’t return anything, we can simply do:

foo?.doSomethingInteresting()

If foo contains a value, this method will be called on it. If it doesn’t, nothing bad will happen – the code will simply continue executing.

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

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