Swift variable decorations with "?" (question mark) and "!" (exclamation mark)

Question

I understand that in Swift all variables must be set with a value, and that by using optionals we can set a variable to be set to nil initially.

What I don't understand is, what setting a variable with a ! is doing, because I was under the impression that this "unwraps" a value from an optional. I thought by doing so, you are guaranteeing that there is a value to unwrap in that variable, which is why on IBActions and such you see it used.

So simply put, what is the variable being initialized to when you do something like this:

var aShape : CAShapeLayer!

And why/when would I do this?

You would do this to tell that a variable is nt nil after you checked this fact. — Matthias
I don't think this should be marked as a duplicate. "What is an optional?" is not the same question as "What is the difference between the two types of optionals?" which is pretty much what this question is — Jiaaro
@Jiaaro even in that case, there are already tons of questions regarding optionals, implicitly unwrapped optionals, and the like. You can also refer to this one: stackoverflow.com/questions/24272781/… — Jack
@JackWu Ok, but I'm pretty sure this question wasn't a duplicate when it was asked. (it was asked a full week before your example, for instance) — Jiaaro
@Jiaaro You make a good point, I didn't notice this was older..maybe that other one should be marked as a duplicate of this one instead.. — Jack

Jiaaro Jiaaro · Accepted Answer · 2014-06-09T14:42:50

In a type declaration the ! is similar to the ?. Both are an optional, but the ! is an "implicitly unwrapped" optional, meaning that you do not have to unwrap it to access the value (but it can still be nil).

This is basically the behavior we already had in objective-c. A value can be nil, and you have to check for it, but you can also just access the value directly as if it wasn't an optional (with the important difference that if you don't check for nil you'll get a runtime error)

// Cannot be nil
var x: Int = 1

// The type here is not "Int", it's "Optional Int"
var y: Int? = 2

// The type here is "Implicitly Unwrapped Optional Int"
var z: Int! = 3

Usage:

// you can add x and z
x + z == 4

// ...but not x and y, because y needs to be unwrapped
x + y // error

// to add x and y you need to do:
x + y!

// but you *should* do this:
if let y_val = y {
    x + y_val
}

Swift variable decorations with "?" (question mark) and "!" (exclamation mark)

1 Answers

Usage: