It’s common in Swift to chain properties and method calls on a variable:

let instrument = "piano"
let firstUppercasedCharacter = instrument.uppercased().first
print(firstUppercasedCharacter) // prints “p”

But what if the variable is an optional? We can use the ! operator to force unwrap the optional:

let instrument: String? = "piano"
let firstUppercasedCharacter = instrument!.uppercased().first
print(firstUppercasedCharacter) // prints p

But this will crash if the optional is nil:

let instrument: String? = nil
let firstUppercasedCharacter = instrument!.uppercased().first // Fatal error: Unexpectedly found nil while unwrapping an Optional value
print(firstUppercasedCharacter)

Fortunately Swift gives us a way to safely chain different calls on an optional using optional chaining. By replacing the ! operator with the ? operator, our code can no longer crash. If the optional is nil, then the entire expression will evaluate to nil.

let instrument: String? = nil
let firstUppercasedCharacter = instrument?.uppercased().first
print(firstUppercasedCharacter) // Prints nil

If the optional is not nil, then it will be unwrapped and have the subsequent properties and methods called on it. Note that because the entire expression can be nil, it now evaluates to an optional as well:

let instrument: String? = "piano"
let firstUppercasedCharacter = instrument?.uppercased().first
print(firstUppercasedCharacter) //Prints “Optional(“P”)"