In this article by Andrew J Wagner, author of the book Learning Swift, we will cover:
What is an optional?
How to unwrap an optional
Optional chaining
Implicitly unwrapped optionals
How to debug optionals
The underlying implementation of an optional
(For more resources related to this topic, see here.)
Introducing optionals
So, we know that the purpose of optionals in Swift is to allow the representation of the absent value, but what does that look like and how does it work? An optional is a special type that can wrap any other type. This means that you can make an optional String, optional Array, and so on. You can do this by adding a question mark (?) to the type name:
var possibleString: String?
var possibleArray: [Int]?
Note that this code does not specify any initial values. This is because all optionals, by default, are set to no value at all. If we want to provide an initial value, we can do so like any other variable:
var possibleInt: Int? = 10
Also note that, if we leave out the type specification (: Int?), possibleInt would be inferred to be of the Int type instead of an Int optional.
It is pretty verbose to say that a variable lacks a value. Instead, if an optional lacks a variable, we say that it is nil. So, both possibleString and possibleArray are nil, while possibleInt is 10. However, possibleInt is not truly 10. It is still wrapped in an optional.
You can see all the forms a variable can take by putting the following code in to a playground:
var actualInt = 10
var possibleInt: Int? = 10
var nilInt: Int?
println(actualInt) // "10"
println(possibleInt) // "Optional(10)"
println(nilInt) // "nil"
As you can see, actualInt prints out as we expect it to, but possibleInt prints out as an optional that contains the value 10 instead of just 10. This is a very important distinction because an optional cannot be used as if it were the value it wraps. The nilInt optional just reports that it is nil. At any point, you can update the value within an optional, including the fact that you can give it a value for the first time using the assignment operator (=):
nilInt = 2
println(nilInt) // "Optional(2)"
You can even remove the value within an optional by assigning it to nil:
nilInt = nil
println(nilInt) // "nil"
So, we have this wrapped form of a variable that may or may not contain a value. What do we do if we need to access the value within an optional? The answer is that we must unwrap it.
Unwrapping an optional
There are multiple ways to unwrap an optional. All of them essentially assert that there is truly a value within the optional. This is a wonderful safety feature of Swift. The compiler forces you to consider the possibility that an optional lacks any value at all. In other languages, this is a very commonly overlooked scenario that can cause obscure bugs.
Optional binding
The safest way to unwrap an optional is using something called optional binding. With this technique, you can assign a temporary constant or variable to the value contained within the optional. This process is contained within an if statement, so that you can use an else statement for when there is no value. An optional binding looks like this:
if let string = possibleString {
println("possibleString has a value: \(string)")
}
else {
println("possibleString has no value")
}
An optional binding is distinguished from an if statement primarily by the if let syntax. Semantically, this code says "if you can let the constant string be equal to the value within possibleString, print out its value; otherwise, print that it has no value." The primary purpose of an optional binding is to create a temporary constant that is the normal (nonoptional) version of the optional.
It is also possible to use a temporary variable in an optional binding:
possibleInt = 10
if var int = possibleInt {
int *= 2
}
println(possibleInt) // Optional(10)
Note that an astrix (*) is used for multiplication in Swift. You should also note something important about this code, that is, if you put it into a playground, even though we multiplied int by 2, the value does not change. When we print out possibleInt later, the value still remains Optional(10). This is because even though we made the int variable (otherwise known as mutable), it is simply a temporary copy of the value within possibleInt. No matter what we do with int, nothing will be changed about the value within possibleInt. If we need to update the actual value stored within possibleInt, we need to simply assign possibleInt to int after we are done modifying it:
possibleInt = 10
if var int = possibleInt {
int *= 2
possibleInt = int
}
println(possibleInt) // Optional(20)
Now the value wrapped inside possibleInt has actually been updated.
A common scenario that you will probably come across is the need to unwrap multiple optional values. One way of doing this is by simply nesting the optional bindings:
if let actualString = possibleString {
if let actualArray = possibleArray {
if let actualInt = possibleInt {
println(actualString)
println(actualArray)
println(actualInt)
}
}
}
However, this can be a pain as it increases the indentation level each time to keep the code organized. Instead, you can actually list multiple optional bindings in a single statement separated by commas:
if let actualString = possibleString,
let actualArray = possibleArray,
let actualInt = possibleInt
{
println(actualString)
println(actualArray)
println(actualInt)
}
This generally produces more readable code.
This way of unwrapping is great, but saying that optional binding is the safe way to access the value within an optional implies that there is an unsafe way to unwrap an optional. This way is called forced unwrapping.
Forced unwrapping
The shortest way to unwrap an optional is by forced unwrapping. This is done using an exclamation mark (!) after the variable name when it is used:
possibleInt = 10
possibleInt! *= 2
println(possibleInt) // "Optional(20)"
However, the reason it is considered unsafe is that your entire program crashes if you try to unwrap an optional that is currently nil:
nilInt! *= 2 // fatal error
The full error you get is "unexpectedly found as nil while unwrapping an optional value". This is because forced unwrapping is essentially your personal guarantee that the optional truly holds a value. This is why it is called forced.
Therefore, forced unwrapping should be used in limited circumstances. It should never be used just to shorten up the code. Instead, it should only be used when you can guarantee, from the structure of the code, that it cannot be nil, even though it is defined as an optional. Even in this case, you should check whether it is possible to use a nonoptional variable instead. The only other place you may use it is when your program truly cannot recover if an optional is nil. In these circumstances, you should at least consider presenting an error to the user, which is always better than simply having your program crash.
An example of a scenario where forced unwrapping may be used effectively is with lazily calculated values. A lazily calculated value is a value that is not created until the first time it is accessed. To illustrate this, let's consider a hypothetical class that represents a filesystem directory. It would have a property that lists its contents that are lazily calculated. The code would look something like this:
class FileSystemItem {}
class File: FileSystemItem {}
class Directory: FileSystemItem {
private var realContents: [FileSystemItem]?
var contents: [FileSystemItem] {
if self.realContents == nil {
self.realContents = self.loadContents()
}
return self.realContents!
}
private func loadContents() -> [FileSystemItem] {
// Do some loading
return []
}
}
Here, we defined a superclass called FileSystemItem that both File and Directory inherit from. The contents of a directory is a list of any kind of FileSystemItem. We define content as a calculated variable and store the real value within the realContents property. The calculated property checks whether there is a value yet loaded for realContents; if there isn't, it loads the contents and puts it into the realContents property. Based on this logic, we know with 100 percent certainty that there will be a value within realContents by the time we get to the return statement, so it is perfectly safe to use forced unwrapping.
Nil coalescing
In addition to optional binding and forced unwrapping, Swift also provides an operator called the nil coalescing operator to unwrap an optional. This is represented by a double question mark (??). Basically, this operator lets us provide a default value for a variable or operation result in case it is nil. This is a safe way to turn an optional value into a nonoptional value and it would look something like this:
var possibleString: String? = "An actual string"
println(possibleString ?? "Default String") // "An Actual String"
Here, we ask the program to print out possibleString unless it is nil, in which case, it will just print Default String. Since we did give it a value, it printed out that value and it is important to note that it printed out as a regular variable, not as an optional. This is because one way or another, an actual value will be printed.
This is a great tool for concisely and safely unwrapping an optional when a default value makes sense.
Optional chaining
A common scenario in Swift is to have an optional that you must calculate something from. If the optional has a value you want to store the result of the calculation on, but if it is nil, the result should just be set to nil:
var invitee: String? = "Sarah"
var uppercaseInvitee: String?
if let actualInvitee = invitee {
uppercaseInvitee = actualInvitee.uppercaseString
}
This is pretty verbose. To shorten this up in an unsafe way, we could use forced unwrapping:
uppercaseInvitee = invitee!.uppercaseString
However, optional chaining will allow us to do this safely. Essentially, it allows optional operations on an optional. When the operation is called, if the optional is nil, it immediately returns nil; otherwise, it returns the result of performing the operation on the value within the optional:
uppercaseInvitee = invitee?.uppercaseString
So in this call, invitee is an optional. Instead of unwrapping it, we will use optional chaining by placing a question mark (?) after it, followed by the optional operation. In this case, we asked for the uppercaseInvitee property on it. If invitee is nil, uppercaseInvitee is immediately set to nil without it even trying to access uppercaseString. If it actually does contain a value, uppercaseInvitee gets set to the uppercaseString property of the contained value. Note that all optional chains return an optional result.
You can chain as many calls, both optional and nonoptional, as you want in this way:
var myNumber: String? = "27"
myNumber?.toInt()?.advancedBy(10).description
This code attempts to add 10 to myNumber, which is represented by String. First, the code uses an optional chain in case myNumber is nil. Then, the call to toInt uses an additional optional chain because that method returns an optional Int type. We then call advancedBy, which does not return an optional, allowing us to access the description of the result without using another optional chain. If at any point any of the optionals are nil, the result will be nil. This can happen for two different reasons:
This can happen because myNumber is nil
This can also happen because toInt returns nil as it cannot convert String to the Int type
If the chain makes it all the way to advanceBy, there is no longer a failure path and it will definitely return an actual value. You will notice that there are exactly two question marks used in this chain and there are two possible failure reasons.
At first, it can be hard to understand when you should and should not use a question mark to create a chain of calls. The rule is that you should always use a question mark if the previous element in the chain returns an optional. However, since you are prepared, let's look at what happens if you use an optional chain improperly:
myNumber.toInt() // Value of optional type 'String?' not unwrapped
In this case, we try to call a method directly on an optional without a chain so that we get an error.
We also have the case where we try to inappropriately use an optional chain:
var otherNumber = "10"
otherNumber?.toInt() // Operand of postfix '?' should have optional type
Here, we get an error that says a question mark can only be used on an optional type. It is great to have a good sense of catching errors, which you will see when you make mistakes, so that you can quickly correct them because we all make silly mistakes from time to time.
Another great feature of optional chaining is that it can be used for method calls on an optional that does not actually return a value:
var invitees: [String]? = []
invitee?.removeAll(keepCapacity: false)
In this case, we only want to call removeAll if there is truly a value within the optional array. So, with this code, if there is a value, all the elements are removed from it: otherwise, it remains nil.
In the end, option chaining is a great choice for writing concise code that still remains expressive and understandable.
Implicitly unwrapped optionals
There is a second type of optional called an implicitly unwrapped optional. There are two ways to look at what an implicitly unwrapped optional is. One way is to say that it is a normal variable that can also be nil. The other way is to say that it is an optional that you don't have to unwrap to use. The important thing to understand about them is that like optionals, they can be nil, but like a normal variable, you do not have to unwrap them.
You can define an implicitly unwrapped optional with an exclamation mark (!) instead of a question mark (?) after the type name:
var name: String!
Just like with regular optionals, implicitly unwrapped optionals do not need to be given an initial value because they are nil by default.
At first, this may sound like it is the best of both worlds, but in reality, it is more like the worst of both worlds. Even though an implicitly unwrapped optional does not have to be unwrapped, it will crash your entire program if it is nil when used:
name.uppercaseString // Crash
A great way to think about them is that every time an implicitly unwrapped optional is used, it is implicitly performing a forced unwrapping. The exclamation mark is placed in its type declaration instead of using it every time. This is particularly bad because it appears the same as any other variable except for how it is declared. This means that it is very unsafe to use, unlike a normal optional.
So, if implicitly unwrapped optionals are the worst of both worlds and are so unsafe, why do they even exist? The reality is that in rare circumstances, they are necessary. They are used in circumstances where a variable is not truly optional, but you also cannot give an initial value to it. This is almost always true in the case of custom types that have a member variable that is nonoptional, but cannot be set during initialization.
A rare example of this is a view in iOS. UIKit, as we discussed earlier, is the framework that Apple provides for iOS development. In it, Apple has a class called UIView that is used for displaying content on the screen. Apple also provides a tool in Xcode called Interface Builder that lets you design these views in a visual editor instead of in code. Many views designed in this way need references to other views that can be accessed programmatically later. When one of these views is loaded, it is initialized without anything connected and then all the connections are made. Once all the connections are made, a function called awakeFromNib is called on the view. This means that these connections are not available for use during initialization, but are available once awakeFromNib is called. This order of operations also ensures that awakeFromNib is always called before anything actually uses the view. This is a circumstance where it is necessary to use an implicitly unwrapped optional. A member variable may not be defined until the view is initialized and when it is completely loaded:
import UIKit class MyView: UIView {
@IBOutlet var button : UIButton!
var buttonOriginalWidth : CGFloat!
override func awakeFromNib() {
self.buttonOriginalWidth = self.button.frame.size.width
}
}
Note that we have actually declared two implicitly unwrapped optionals. The first is a connection to button. We know this is a connection because it is preceded by @IBOutlet. This is declared as an implicitly unwrapped optional because the connections are not set up until after initialization, but they are still guaranteed to be set up before any other methods are called on the view.
This also then leads us to make our second variable, buttonOriginalWidth, implicitly unwrapped because we need to wait until the connection is made before we can determine the width of button. After awakeFromNib is called, it is safe to treat both button and buttonOriginalWidth as nonoptional.
You may have noticed that we had to dive pretty deep in to app development in order to find a valid use case for implicitly unwrapped optionals, and this is arguably only because UIKit is implemented in Objective-C.
Debugging optionals
We already saw a couple of compiler errors that we commonly see because of optionals. If we try to call a method on an optional that we intended to call on the wrapped value, we will get an error. If we try to unwrap a value that is not actually optional, we will get an error that the variable or constant is not optional. We also need to be prepared for runtime errors that optionals can cause.
As discussed, optionals cause runtime errors if you try to forcefully unwrap an optional that is nil. This can happen with both explicit and implicit forced unwrapping. If you followed my advice so far in this article, this should be a rare occurrence. However, we all end up working with third-party code, and maybe they were lazy or maybe they used forced unwrapping to enforce their expectations about how their code should be used.
Also, we all suffer from laziness from time to time. It can be exhausting or discouraging to worry about all the edge cases when you are excited about programming the main functionality of your app. We may use forced unwrapping temporarily while we worry about that main functionality and plan to come back to handle it later. After all, during development, it is better to have a forced unwrapping crash the development version of your app than it is for it to fail silently if you have not yet handled that edge case. We may even decide that an edge case is not worth the development effort of handling because everything about developing an app is a trade-off. Either way, we need to recognize a crash from forced unwrapping quickly, so that we don't waste extra time trying to figure out what went wrong.
When an app tries to unwrap a nil value, if you are currently debugging the app, Xcode shows you the line that tries to do the unwrapping. The line reports that there was EXC_BAD_INSTRUCTION and you will also get a message in the console saying fatal error: unexpectedly found nil while unwrapping an Optional value:
You will also sometimes have to look at which code currently calls the code that failed. To do that, you can use the call stack in Xcode. When your program crashes, Xcode automatically displays the call stack, but you can also manually show it by going to View | Navigators | Show Debug Navigator. This will look something as follows:
Here, you can click on different levels of code to see the state of things. This becomes even more important if the program crashes within one of Apple's framework, where you do not have access to the code. In that case, you should move up the call stack to the point where your code is called in the framework. You may also be able to look at the names of the functions to help you figure out what may have gone wrong.
Anywhere on the call stack, you can look at the state of the variables in the debugger, as shown in the following screenshot:
If you do not see this variable's view, you can display it by clicking on the button at the bottom-left corner, which is second from the right that will be grayed out. Here, you can see that invitee is indeed nil, which is what caused the crash.
As powerful as the debugger is, if you find that it isn't helping you find the problem, you can always put println statements in important parts of the code. It is always safe to print out an optional as long as you don't forcefully unwrap it like in the preceding example. As we saw earlier, when an optional is printed, it will print nil if it doesn't have a value or it will print Optional(<value>) if it does have a value.
Debugging is an extremely important part of becoming a productive developer because we all make mistakes and create bugs. Being a great developer means that you can identify problems quickly and understand how to fix them soon after that. This will largely come from practice, but it will also come when you have a firm grasp of what really happens with your code instead of simply adapting some code you find online to fit your needs through trial and error.
The underlying implementation
At this point, you should have a pretty strong grasp of what an optional is and how to use and debug it, but it is valuable to look deeper at optionals and see how they actually work.
In reality, the question mark syntax for optionals is just a special shorthand. Writing String? is equivalent to writing Optional<String>. Writing String! is equivalent to writing ImplicitlyUnwrappedOptional<String>. The Swift compiler has shorthand versions because they are so commonly used This allows the code to be more concise and readable.
If you declare an optional using the long form, you can see Swift's implementation by holding command and clicking on the word Optional. Here, you can see that Optional is implemented as an enumeration. If we simplify the code a little, we have:
enum Optional<T> {
case None
case Some(T)
}
So, we can see that Optional really has two cases: None and Some. None stands for the nil case, while the Some case has an associated value, which is the value wrapped inside Optional. Unwrapping is then the process of retrieving the associated value out of the Some case.
One part of this that you have not seen yet is the angled bracket syntax (<T>). This is a generic and essentially allows the enumeration to have an associated value of any type.
Realizing that optionals are simply enumerations will help you to understand how to use them. It also gives you some insight into how concepts are built on top of other concepts. Optionals seem really complex until you realize that they are just two-case enumerations. Once you understand enumerations, you can pretty easily understand optionals as well.
Summary
We only covered a single concept, optionals, in this article, but we saw that this is a pretty dense topic. We saw that at the surface level, optionals are pretty straightforward. They offer a way to represent a variable that has no value. However, there are multiple ways to get access to the value wrapped within an optional, which have very specific use cases. Optional binding is always preferred as it is the safest method, but we can also use forced unwrapping if we are confident that an optional is not nil. We also have a type called implicitly unwrapped optional to delay the assigning of a variable that is not intended to be optional, but we should use it sparingly because there is almost always a better alternative.
Resources for Article:
Further resources on this subject:
Network Development with Swift [article]
Flappy Swift [article]
Playing with Swift [article]
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