How to deal with "Fatal error: unexpectedly found nil while unwrapping an Optional value." [duplicate]

My Swift program is crashing with EXC_BAD_INSTRUCTION and one of the following similar errors. What does this error mean, and how do I fix it?

Fatal error: Unexpectedly found nil while unwrapping an Optional value

or

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

^{This post is intended to collect answers to "unexpectedly found nil" issues, so that they are not scattered and hard to find. Feel free to add your own answer or edit the existing wiki answer.}

Solution 1:

^{This answer is community wiki. If you feel it could be made better, feel free to edit it!}

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.

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.

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.

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)}

This can therefore also be used to set properties as well as call methods. For example:

foo?.bar = Bar()

Again, nothing bad will happen here if foo is nil. Your code will simply continue executing.

Another neat trick that optional chaining lets you do is check whether setting a property or calling a method was successful. You can do this by comparing the return value to nil.

^{(This is because an optional value will return Void? rather than Void on a method that doesn’t return anything)}

For example:

if (foo?.bar = Bar()) != nil {
    print("bar was set successfully")
} else {
    print("bar wasn’t set successfully")
}

However, things become a little bit more tricky when trying to access properties or call methods that return a value. Because foo is optional, anything returned from it will also be optional. To deal with this, you can either unwrap the optionals that get returned using one of the above methods – or unwrap foo itself before accessing methods or calling methods that return values.

Also, as the name suggests, you can ‘chain’ these statements together. This means that if foo has an optional property baz, which has a property qux – you could write the following:

let optionalQux = foo?.baz?.qux

Again, because foo and baz are optional, the value returned from qux will always be an optional regardless of whether qux itself is optional.

map and flatMap

An often underused feature with optionals is the ability to use the map and flatMap functions. These allow you to apply non-optional transforms to optional variables. If an optional has a value, you can apply a given transformation to it. If it doesn’t have a value, it will remain nil.

For example, let’s say you have an optional string:

let anOptionalString:String?

By applying the map function to it – we can use the stringByAppendingString function in order to concatenate it to another string.

Because stringByAppendingString takes a non-optional string argument, we cannot input our optional string directly. However, by using map, we can use allow stringByAppendingString to be used if anOptionalString has a value.

For example:

var anOptionalString:String? = "bar"

anOptionalString = anOptionalString.map {unwrappedString in
    return "foo".stringByAppendingString(unwrappedString)
}

print(anOptionalString) // Optional("foobar")

However, if anOptionalString doesn’t have a value, map will return nil. For example:

var anOptionalString:String?

anOptionalString = anOptionalString.map {unwrappedString in
    return "foo".stringByAppendingString(unwrappedString)
}

print(anOptionalString) // nil

flatMap works similarly to map, except it allows you to return another optional from within the closure body. This means you can input an optional into a process that requires a non-optional input, but can output an optional itself.

try!

Swift's error handling system can be safely used with Do-Try-Catch:

do {
    let result = try someThrowingFunc() 
} catch {
    print(error)
}

If someThrowingFunc() throws an error, the error will be safely caught in the catch block.

The error constant you see in the catch block has not been declared by us - it's automatically generated by catch.

You can also declare error yourself, it has the advantage of being able to cast it to a useful format, for example:

do {
    let result = try someThrowingFunc()    
} catch let error as NSError {
    print(error.debugDescription)
}

Using try this way is the proper way to try, catch and handle errors coming from throwing functions.

There's also try? which absorbs the error:

if let result = try? someThrowingFunc() {
    // cool
} else {
    // handle the failure, but there's no error information available
}

But Swift's error handling system also provides a way to "force try" with try!:

let result = try! someThrowingFunc()

The concepts explained in this post also apply here: if an error is thrown, the application will crash.

You should only ever use try! if you can prove that its result will never fail in your context - and this is very rare.

Most of the time you will use the complete Do-Try-Catch system - and the optional one, try?, in the rare cases where handling the error is not important.

Resources

• Apple documentation on Swift Optionals
• When to use and when not to use implicitly unwrapped optionals
• Learn how to debug an iOS app crash

Solution 2:

TL;DR answer

With very few exceptions, this rule is golden:

Avoid use of !

Declare variable optional (?), not implicitly unwrapped optionals (IUO) (!)

In other words, rather use:
var nameOfDaughter: String?

Instead of:
var nameOfDaughter: String!

Unwrap optional variable using if let or guard let

Either unwrap variable like this:

if let nameOfDaughter = nameOfDaughter {
    print("My daughters name is: \(nameOfDaughter)")
}

Or like this:

guard let nameOfDaughter = nameOfDaughter else { return }
print("My daughters name is: \(nameOfDaughter)")

This answer was intended to be concise, for full comprehension read accepted answer

Resources

• Avoiding force unwrapping

Solution 3:

This question comes up ALL THE TIME on SO. It's one of the first things that new Swift developers struggle with.

Background:

Swift uses the concept of "Optionals" to deal with values that could contain a value, or not. In other languages like C, you might store a value of 0 in a variable to indicate that it contains no value. However, what if 0 is a valid value? Then you might use -1. What if -1 is a valid value? And so on.

Swift optionals let you set up a variable of any type to contain either a valid value, or no value.

You put a question mark after the type when you declare a variable to mean (type x, or no value).

An optional is actually a container than contains either a variable of a given type, or nothing.

An optional needs to be "unwrapped" in order to fetch the value inside.

The "!" operator is a "force unwrap" operator. It says "trust me. I know what I am doing. I guarantee that when this code runs, the variable will not contain nil." If you are wrong, you crash.

Unless you really do know what you are doing, avoid the "!" force unwrap operator. It is probably the largest source of crashes for beginning Swift programmers.

How to deal with optionals:

There are lots of other ways of dealing with optionals that are safer. Here are some (not an exhaustive list)

You can use "optional binding" or "if let" to say "if this optional contains a value, save that value into a new, non-optional variable. If the optional does not contain a value, skip the body of this if statement".

Here is an example of optional binding with our foo optional:

if let newFoo = foo //If let is called optional binding. {
  print("foo is not nil")
} else {
  print("foo is nil")
}

Note that the variable you define when you use optional biding only exists (is only "in scope") in the body of the if statement.

Alternately, you could use a guard statement, which lets you exit your function if the variable is nil:

func aFunc(foo: Int?) {
  guard let newFoo = input else { return }
  //For the rest of the function newFoo is a non-optional var
}

Guard statements were added in Swift 2. Guard lets you preserve the "golden path" through your code, and avoid ever-increasing levels of nested ifs that sometimes result from using "if let" optional binding.

There is also a construct called the "nil coalescing operator". It takes the form "optional_var ?? replacement_val". It returns a non-optional variable with the same type as the data contained in the optional. If the optional contains nil, it returns the value of the expression after the "??" symbol.

So you could use code like this:

let newFoo = foo ?? "nil" // "??" is the nil coalescing operator
print("foo = \(newFoo)")

You could also use try/catch or guard error handling, but generally one of the other techniques above is cleaner.

EDIT:

Another, slightly more subtle gotcha with optionals is "implicitly unwrapped optionals. When we declare foo, we could say:

var foo: String!

In that case foo is still an optional, but you don't have to unwrap it to reference it. That means any time you try to reference foo, you crash if it's nil.

So this code:

var foo: String!


let upperFoo = foo.capitalizedString

Will crash on reference to foo's capitalizedString property even though we're not force-unwrapping foo. the print looks fine, but it's not.

Thus you want to be really careful with implicitly unwrapped optionals. (and perhaps even avoid them completely until you have a solid understanding of optionals.)

Bottom line: When you are first learning Swift, pretend the "!" character is not part of the language. It's likely to get you into trouble.

Solution 4:

Since the above answers clearly explains how to play safely with Optionals. I will try explain what Optionals are really in swift.

Another way to declare an optional variable is

var i : Optional<Int>

And Optional type is nothing but an enumeration with two cases, i.e

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

So to assign a nil to our variable 'i'. We can do var i = Optional<Int>.none or to assign a value, we will pass some value var i = Optional<Int>.some(28)

According to swift, 'nil' is the absence of value. And to create an instance initialized with nil We have to conform to a protocol called ExpressibleByNilLiteral and great if you guessed it, only Optionals conform to ExpressibleByNilLiteral and conforming to other types is discouraged.

ExpressibleByNilLiteral has a single method called init(nilLiteral:) which initializes an instace with nil. You usually wont call this method and according to swift documentation it is discouraged to call this initializer directly as the compiler calls it whenever you initialize an Optional type with nil literal.

Even myself has to wrap (no pun intended) my head around Optionals :D Happy Swfting All.

Solution 5:

First, you should know what an Optional value is. You can step to The Swift Programming Language for detail.

Second, you should know the optional value has two statuses. One is the full value, and the other is a nil value. So before you implement an optional value, you should check which state it is.

You can use if let ... or guard let ... else and so on.

One other way, if you don't want to check the variable state before your implementation, you can also use var buildingName = buildingName ?? "buildingName" instead.