“致命错误：在展开可选值时意外发现 nil”是什么意思？

Question

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.}

Answer 1

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 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

Answer 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

Answer 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.

Answer 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, ie

 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 .

Answer 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" var buildingName = buildingName ?? "buildingName" instead.

Answer 6

I had this error once when I was trying to set my Outlets values from the prepare for segue method as follows:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let destination = segue.destination as? DestinationVC{

        if let item = sender as? DataItem{
            // This line pops up the error
            destination.nameLabel.text = item.name
        }
    }
}

Then I found out that I can't set the values of the destination controller outlets because the controller hasn't been loaded or initialized yet.

So I solved it this way:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let destination = segue.destination as? DestinationVC{

        if let item = sender as? DataItem{
            // Created this method in the destination Controller to update its outlets after it's being initialized and loaded
            destination.updateView(itemData:  item)
        }
    }
}

Destination Controller:

// This variable to hold the data received to update the Label text after the VIEW DID LOAD
var name = ""

// Outlets
@IBOutlet weak var nameLabel: UILabel!

override func viewDidLoad() {
    super.viewDidLoad()

    // Do any additional setup after loading the view.
    nameLabel.text = name
}

func updateView(itemDate: ObjectModel) {
    name = itemDate.name
}

I hope this answer helps anyone out there with the same issue as I found the marked answer is great resource to the understanding of optionals and how they work but hasn't addressed the issue itself directly.

Answer 7

Basically you tried to use a nil value in places where Swift allows only non-nil ones, by telling the compiler to trust you that there will never be nil value there, thus allowing your app to compile.

There are several scenarios that lead to this kind of fatal error:

1. forced unwraps:

    let user = someVariable!

   If someVariable is nil, then you'll get a crash. By doing a force unwrap you moved the nil check responsibility from the compiler to you, basically by doing a forced unwrap you're guaranteeing to the compiler that you'll never have nil values there. And guess what it happens if somehow a nil value ends in in someVariable ?

   Solution? Use optional binding (aka if-let), do the variable processing there:

    if user = someVariable { // do your stuff }

2. forced (down)casts:

    let myRectangle = someShape as! Rectangle

   Here by force casting you tell the compiler to no longer worry, as you'll always have a Rectangle instance there. And as long as that holds, you don't have to worry. The problems start when you or your colleagues from the project start circulating non-rectangle values.

   Solution? Use optional binding (aka if-let), do the variable processing there:

    if let myRectangle = someShape as? Rectangle { // yay, I have a rectangle }

3. Implicitly unwrapped optionals. Let's assume you have the following class definition:

    class User { var name: String! init() { name = "(unnamed)" } func nicerName() { return "Mr/Ms " + name } }

   Now, if no-one messes up with the name property by setting it to nil , then it works as expected, however if User is initialized from a JSON that lacks the name key, then you get the fatal error when trying to use the property.

   Solution? Don't use them :) Unless you're 102% sure that the property will always have a non-nil value by the time it needs to be used. In most cases converting to an optional or non-optional will work. Making it non-optional will also result in the compiler helping you by telling the code paths you missed giving a value to that property

4. Unconnected, or not yet connected, outlets. This is a particular case of scenario #3. Basically you have some XIB-loaded class that you want to use.

    class SignInViewController: UIViewController { @IBOutlet var emailTextField: UITextField! }

   Now if you missed connecting the outlet from the XIB editor, then the app will crash as soon as you'll want to use the outlet. Solution? Make sure all outlets are connected. Or use the ? operator on them: emailTextField?.text = "my@email.com" . Or declare the outlet as optional, though in this case the compiler will force you to unwrap it all over the code.

5. Values coming from Objective-C, and that don't have nullability annotations. Let's assume we have the following Objective-C class:

    @interface MyUser: NSObject @property NSString *name; @end

   Now if no nullability annotations are specified (either explicitly or via NS_ASSUME_NONNULL_BEGIN / NS_ASSUME_NONNULL_END ), then the name property will be imported in Swift as String! (an IUO - implicitly unwrapped optional). As soon as some swift code will want to use the value, it will crash if name is nil.

   Solution? Add nullability annotations to your Objective-C code. Beware though, the Objective-C compiler is a little bit permissive when it comes to nullability, you might end up with nil values, even if you explicitly marked them as nonnull .

Answer 8

This is more of a important comment and that why implicitly unwrapped optionals can be deceptive when it comes to debugging nil values.

Think of the following code: It compiles with no errors/warnings:

c1.address.city = c3.address.city

Yet at runtime it gives the following error: Fatal error: Unexpectedly found nil while unwrapping an Optional value

Can you tell me which object is nil ?

You can't!

The full code would be:

class ViewController: UIViewController {

    override func viewDidLoad() {
        super.viewDidLoad()

        var c1 = NormalContact()
        let c3 = BadContact()

        c1.address.city = c3.address.city // compiler hides the truth from you and then you sudden get a crash
    }
}

struct NormalContact {
    var address : Address = Address(city: "defaultCity")
}

struct BadContact {
    var address : Address!
}

struct Address {
    var city : String
}

Long story short by using var address : Address! you're hiding the possibility that a variable can be nil from other readers. And when it crashes you're like "what the hell?! my address isn't an optional, so why am I crashing?!.

Hence it's better to write as such:

c1.address.city = c2.address!.city  // ERROR:  Fatal error: Unexpectedly found nil while unwrapping an Optional value 

Can you now tell me which object it is that was nil ?

This time the code has been made more clear to you. You can rationalize and think that likely it's the address parameter that was forcefully unwrapped.

The full code would be :

class ViewController: UIViewController {

    override func viewDidLoad() {
        super.viewDidLoad()

        var c1 = NormalContact()
        let c2 = GoodContact()

        c1.address.city = c2.address!.city
        c1.address.city = c2.address?.city // not compile-able. No deceiving by the compiler
        c1.address.city = c2.address.city // not compile-able. No deceiving by the compiler
        if let city = c2.address?.city {  // safest approach. But that's not what I'm talking about here. 
            c1.address.city = city
        }

    }
}

struct NormalContact {
    var address : Address = Address(city: "defaultCity")
}

struct GoodContact {
    var address : Address?
}

struct Address {
    var city : String
}

Answer 9

The errors EXC_BAD_INSTRUCTION and fatal error: unexpectedly found nil while implicitly unwrapping an Optional value appears the most when you have declared an @IBOutlet , but not connected to the storyboard .

You should also learn about how Optionals work, mentioned in other answers, but this is the only time that mostly appears to me.

Answer 10

If you get this error in CollectionView try to create CustomCell file and Custom xib also.

add this code in ViewDidLoad() at mainVC.

    let nib = UINib(nibName: "CustomnibName", bundle: nil)
    self.collectionView.register(nib, forCellWithReuseIdentifier: "cell")

Answer 11

Xcode 12 iOS 14 Swift 5

My problem was the type of navigation as I called the vie controller direct without instantiating the storyboard so that's mean data was not set yet from the storyboard.

When you navigate, navigate with

let homeViewController = UIStoryboard(name: "Main", bundle: nil).instantiateViewController(withIdentifier: "home") as? HomeEventsViewController
    homeViewController?.modalTransitionStyle = .crossDissolve
    homeViewController?.modalPresentationStyle = .fullScreen
    view.present(homeViewController ?? UIViewController(), animated: true, completion: nil)

Hopefully it works :-)

Answer 12

Swift 5.7 +

if let shorthand for shadowing an existing optional variable

Above answers clearly explains why this issue arises and how to handle this issue. But there is a new way of handling this issue from swift 5.7+ onwards.

var myVariable: Int?

Previously

if let myVariable = myVariable {
    //this part get executed if the variable is not nil
}else{
    //this part get executed if the variable is nil
}

now

here now we can omit the right hand side of the expression.

if let myVariable {
    //this part get executed if the variable is not nil
}else{
    //this part get executed if the variable is nil
}

Previously we had to repeat the referenced identifier twice, which can cause these optional binding conditions to be verbose, especially when using lengthy variable names.

But now there is a shorthand syntax for optional binding by omitting the right hand side of the expression.

Same thing is applicable for guard let statement.

For more details :

Proposal for if-let shorthand

Answer 13

I came across this error while making a segue from a table view controller to a view controller because I had forgotten to specify the custom class name for the view controller in the main storyboard.

Something simple that is worth checking if all else looks ok

Answer 14

If my case I set a variable to UILabel which was nil.

So I fixed it and thereafter it did not throw the error.

Code snippet

class ResultViewController: UIViewController {

    @IBOutlet weak var resultLabel: UILabel!
    var bmiValue=""
    override func viewDidLoad() {
        super.viewDidLoad()
        print(bmiValue)
        resultLabel.text=bmiValue //where bmiValue was nil , I fixed it and problem was solved

    }
    
    @IBAction func recaculateBmi(_ sender: UIButton) {
        self.dismiss(animated: true, completion: nil)
    }
    

}

Answer 15

in simple words you are trying to use a value of the optional variable which is nil. quick fix could be use guard or if let instead of force unwrap like putting ! at the end of variable

Answer 16

这个问题可能是重复的，但这是因为它是一个常见问题，但这里有一篇完整的博客文章，供您全面理解。

Answer 17

花一些时间，可能对其他人有用：插座对我来说是零（因此抛出错误：在展开一个Optional值时意外发现nil）是因为： IBOutlet是nil，但是它连接到情节提要中，Swift

Answer 18

This is because you are trying to use a value which can possible be nil, but you decided you don't want to have to check it, but instead assume its set when you uses it and define it as !, there are different philosophies on use of variable set as force unwrap, some people are against there use at all, I personal think they are ok for things that will crash all the time and are simple to reason about, usually references to resource, like outlets to xib files, or uses of images with you app that are part of your assets, if these are not set up properly, you app is going to crash straight away, for a very obvious reason, you can get into difficult when the order of objects being created can be uncertain, and trying to reason solutions to this can be difficult, it usually means a bad design as even it you make them optional, calls to you optional variable may not ever be executed, some projects can demand use of force unwraps for security reasons, things like banking apps, because they want the app to crash rather then continue to work in an unplanned way.

Answer 19

Swift 5 and above ( optionals and optional chaining )

class Address {
    var buildingName: String?
    var buildingNumber: String?
    var street: String?
    func buildingIdentifier() -> String? {
        if let buildingNumber = buildingNumber, let street = street {
            return "\(buildingNumber) \(street)"
        } else if buildingName != nil {
            return buildingName
        } else {
            return nil
        }
    }
}