Is It Safe to Force Unwrap an Optional After Checking It Is Not Nil

Is it safe to force unwrap an optional after checking it is not nil

The difference between if and guard is just simple, but in your case, you should use if, not guard. guard should be used when some values are expected to be present for the function to execute as intended. You don't want error to be present, but if it is not nil, you have to return. So you should use if let:

if let error = error {
    print(error)
    return
}

This way, you don't need to use force unwrapping and it improves how your code is represented.

Back to your question "In general, is this force unwrapping thread safe or not?", assuming that you are really talking about thread safety, it is certainly thread safe. Closure variables are immutable.

If you are talking about safety (ie will "fatal error: unexpectedly found nil while unwrapping an Optional value" occur), it is also certainly safe. error must not be nil before you can unwrap it (because of guard), so the force unwrapping is safe.

In conclusion, your current code is bulletproof, but using if let might be better.

force unwrapping after checking if nil?

Yes, it is bad practice. Perhaps not in this case specifically but it is possible that some other thread could update the property and make it nil between this thread checking for nil and then force-unwrapping.

Just do:

if let arrayOfStrings = self.arrayOfStrings {
    textLabel.text = arrayOfStrings[0]
}

There is no unnecessary variable here since you actually use the variable inside the if let.

Of course in this very specific case of trying to get the first value of an optional array, you can simply do:

textLabel.text = self.arrayOfStrings?.first

If you actually wanted something other than index 0, you should check the index:

if let arrayOfStrings = self.arrayOfStrings, someIndex < arrayOfStrings.count {
    textLabel.text = arrayOfStrings[someIndex]
}

In none of these cases is the "extra" variable a waste of effort or memory. It is being used in a read-only capacity as a constant and Swift is smart enough not to create a complete copy of the array in the process.

Swift why unwrapping optional by guarding against NIL is not working

The type of s is still Optional, so whether you did a nil check or not is irrelevant. The nil check is done in runtime, while the type system is doing a compile-time check. The only way to ensure s can never be nil is via optional binding if let or guard let.

Is it safe to force unwrap variables that have been optionally accessed in the same line of code?

Optional Chaining
from "The Swift Programming Language"
gives the following example:

 let john = Person()
 // ...
 let someAddress = Address()
 // ...
 john.residence?.address = someAddress

followed by (emphasis added):

In this example, the attempt to set the address property of john.residence will fail, because john.residence is currently nil.

The assignment is part of the optional chaining, which means none of the code on the right hand side of the = operator is evaluated.

Applied to your case: In

self?.variable = self!.otherVariable

the right-hand side is not evaluated if self is nil.
Therefore the answer to your question

If self indeed is nil, the second part will never happen?

is "yes". With regard to the second question

And it will never happen that self could be 'nilled' during this single line of code?

My original assumption was that once self has been determined to be != nil,
a strong reference to self! is held throughout the evaluation of the
statement, so that this can not happen. However (as @Hamish pointed out),
this is not guaranteed. Apple engineer Joe Groff writes at Confirming order of operations in
the Swift forum:

This isn't guaranteed. Releases may be optimized to happen earlier than this, to any point after the last formal use of the strong reference. Since the strong reference loaded in order to evaluate the left-hand side weakProperty?.variable is not used afterward, there is nothing keeping it alive, so it could be immediately released.

If there are any side effects in the getter for variable that cause the object referenced by weakProperty to be deallocated, nil-ing out the weak reference, then that would cause the force-unwrap on the right side to fail.
You should use if let to test the weak reference, and reference the strong reference bound by the if let

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)}

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.

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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

Why create Implicitly Unwrapped Optionals, since that implies you know there's a value?

Consider the case of an object that may have nil properties while it's being constructed and configured, but is immutable and non-nil afterwards (NSImage is often treated this way, though in its case it's still useful to mutate sometimes). Implicitly unwrapped optionals would clean up its code a good deal, with relatively low loss of safety (as long as the one guarantee held, it would be safe).

(Edit) To be clear though: regular optionals are nearly always preferable.

Forced unwrap of a return type?

func myFunction() -> NSData { ... }

The above means that myFunction returns an instance of NSData.

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func myFunction() -> NSData? { ... }

The above means that myFunction returns a value of type Optional<NSData> (also called NSData?). Optional is an enum with two cases: .some(value) and .none (also called nil). So the function returns either an NSData (wrapped in the .some case), or nil.

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func myFunction() -> NSData! { ... }

The above means that myFunction returns a value of type Optional<NSData>, just like the NSData? return type in the previous example.

However, the use of ! means that, if you use the value of myFunction() in a way that doesn't type-check, the compiler will try unwrapping the returned value for you. So if you say this:

let maybeLength = myFunction()?.length

then the compiler will see that you are treating the return value of myFunction as an Optional<NSData>. But if you say this:

let dataLength = myFunction().length

then the compiler will see that Optional<NSData> doesn't have a length member, so it will pretend you wrote this:

let dataLength = myFunction()!.length

and if that compiles, it will carry on.

This behavior was defined in Swift Evolution proposal SE-0054, Abolish ImplicitlyUnwrappedOptional type.

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Now let's finally consider this:

func myFunction() -> NSData! { ... }
let data = myFunction()
data?.write()

What is the type of data? The compiler has to infer the type of data based on the expression on the right side of the =. As I explained in the example above, the type of myFunction() is Optional<NSData>. So the type of data is Optional<NSData>, also called NSData?.

The compiler doesn't run into any trouble making data's type NSData?, so it has no reason to implicitly unwrap the return value of myFunction.

Why does a `nil` implicitly unwrapped optional print `nil` and not crash?

That does not crash because print accepts Any as the first parameter. Are implicitly unwrapped optionals a kind of Any? Yes they are! Anything is Any. There is no need to unwrap the optional. An implicitly unwrapped optional can be used in a place where Any is expected without unwrapping the optional.

That could potentially be confusing because now you have something with type Any, which doesn't look like it's optional, but it is an optional under the hood. To avoid this, Swift will output a warning telling you that you are implicitly coercing whatever optional type to Any.

You need to use ! to force unwrap it here:

print(x!)

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