What is an “unwrapped value” in Swift?

First, you have to understand what an Optional type is. An optional type basically means that the variable can be nil.

Example:

var canBeNil : Int? = 4
canBeNil = nil

The question mark indicates the fact that canBeNil can be nil.

This would not work:

var cantBeNil : Int = 4
cantBeNil = nil // can't do this

To get the value from your variable if it is optional, you have to unwrap it. This just means putting an exclamation point at the end.

var canBeNil : Int? = 4
println(canBeNil!)

Your code should look like this:

let optionalSquare: Square? = Square(sideLength: 2.5, name: "optional square") 
let sideLength = optionalSquare!.sideLength

A sidenote:

You can also declare optionals to automatically unwrap by using an exclamation mark instead of a question mark.

Example:

var canBeNil : Int! = 4
print(canBeNil) // no unwrapping needed

So an alternative way to fix your code is:

let optionalSquare: Square! = Square(sideLength: 2.5, name: "optional square") 
let sideLength = optionalSquare.sideLength

EDIT:

The difference that you’re seeing is exactly the symptom of the fact that the optional value is wrapped. There is another layer on top of it. The unwrapped version just shows the straight object because it is, well, unwrapped.

A quick playground comparison:

In the first and second cases, the object is not being automatically unwrapped, so you see two “layers” ({{...}}), whereas in the third case, you see only one layer ({...}) because the object is being automatically unwrapped.

The difference between the first case and the second two cases is that the second two cases will give you a runtime error if optionalSquare is set to nil. Using the syntax in the first case, you can do something like this:

if let sideLength = optionalSquare?.sideLength {
    println("sideLength is not nil")
} else {
    println("sidelength is nil")
}