Swift – What does an exclamation mark mean in the Swift language

forced-unwrappingoptionalswift

The Swift Programming Language guide has the following example:

class Person {
    let name: String
    init(name: String) { self.name = name }
    var apartment: Apartment?
    deinit { println("\(name) is being deinitialized") }
}

class Apartment {
    let number: Int
    init(number: Int) { self.number = number }
    var tenant: Person?
    deinit { println("Apartment #\(number) is being deinitialized") }
}

var john: Person?
var number73: Apartment?

john = Person(name: "John Appleseed")
number73 = Apartment(number: 73)

//From Apple's “The Swift Programming Language” guide (https://developer.apple.com/library/content/documentation/Swift/Conceptual/Swift_Programming_Language/AutomaticReferenceCounting.html)

Then when assigning the apartment to the person, they use an exclamation point to "unwrap the instance":

john!.apartment = number73

What does it mean to "unwrap the instance"? Why is it necessary? How is it different from just doing the following:

john.apartment = number73

I'm very new to the Swift language. Just trying to get the basics down.

UPDATE:
The big piece of the puzzle that I was missing (not directly stated in the answers – at least not at the time of writing this) is that when you do the following:

var john: Person?

that does NOT mean that "john is of type Person and it might be nil", as I originally thought. I was simply misunderstanding that Person and Person? are completely separate types. Once I grasped that, all of the other ?, ! madness, and the great answers below, made a lot more sense.

Best Solution

What does it mean to "unwrap the instance"? Why is it necessary?

As far as I can work out (this is very new to me, too)...

The term "wrapped" implies we should think of an Optional variable as a present, wrapped in shiny paper, which might (sadly!) be empty.

When "wrapped", the value of an Optional variable is an enum with two possible values (a little like a Boolean). This enum describes whether the variable holds a value (Some(T)), or not (None).

If there is a value, this can be obtained by "unwrapping" the variable (obtaining the T from Some(T)).

How is john!.apartment = number73 different from john.apartment = number73? (Paraphrased)

If you write the name of an Optional variable (eg text john, without the !), this refers to the "wrapped" enum (Some/None), not the value itself (T). So john isn't an instance of Person, and it doesn't have an apartment member:

john.apartment
// 'Person?' does not have a member named 'apartment'

The actual Person value can be unwrapped in various ways:

  • "forced unwrapping": john! (gives the Person value if it exists, runtime error if it is nil)
  • "optional binding": if let p = john { println(p) } (executes the println if the value exists)
  • "optional chaining": john?.learnAboutSwift() (executes this made-up method if the value exists)

I guess you choose one of these ways to unwrap, depending upon what should happen in the nil case, and how likely that is. This language design forces the nil case to be handled explicitly, which I suppose improves safety over Obj-C (where it is easy to forget to handle the nil case).

Update:

The exclamation mark is also used in the syntax for declaring "Implicitly Unwrapped Optionals".

In the examples so far, the john variable has been declared as var john:Person?, and it is an Optional. If you want the actual value of that variable, you must unwrap it, using one of the three methods above.

If it were declared as var john:Person! instead, the variable would be an Implicitly Unwrapped Optional (see the section with this heading in Apple's book). There is no need to unwrap this kind of variable when accessing the value, and john can be used without additional syntax. But Apple's book says:

Implicitly unwrapped optionals should not be used when there is a possibility of a variable becoming nil at a later point. Always use a normal optional type if you need to check for a nil value during the lifetime of a variable.

Update 2:

The article "Interesting Swift Features" by Mike Ash gives some motivation for optional types. I think it is great, clear writing.

Update 3:

Another useful article about the implicitly unwrapped optional use for the exclamation mark: "Swift and the Last Mile" by Chris Adamson. The article explains that this is a pragmatic measure by Apple used to declare the types used by their Objective-C frameworks which might contain nil. Declaring a type as optional (using ?) or implicitly unwrapped (using !) is "a tradeoff between safety and convenience". In the examples given in the article, Apple have chosen to declare the types as implicitly unwrapped, making the calling code more convenient, but less safe.

Perhaps Apple might comb through their frameworks in the future, removing the uncertainty of implicitly unwrapped ("probably never nil") parameters and replacing them with optional ("certainly could be nil in particular [hopefully, documented!] circumstances") or standard non-optional ("is never nil") declarations, based on the exact behaviour of their Objective-C code.