Accessing valueless implicitly unwrapped optional?

1
votes

I am curious about the way implicitly unwrapped optionals work, in the Apple reference book - The Swift Programming Language it states:

“If you try to access an implicitly unwrapped optional when it does not contain a value, you will trigger a runtime error. The result is exactly the same as if you place an exclamation mark after a normal optional that does not contain a value.”

If I try and access a valueless optional without an exclamation mark you get an error as stated above.

var optionalVar:String?
println("VAR: \(optionalVar!)")

If however you try an access an implicitly unwrapped optional that does not contain a value the println outputs VAR: nil which is not what is stated above.

var implicitOpt:String!
println("VAR: \(implicitOpt)")

Can anyone clarify this, is it badly worded docs, me miss understanding, a bug?

2
iosswiftoptional

2 Answers

3
votes

Thing is that you are not unwrapping the value in this case, but printing unwrapped optional as special type. Weird, I know. Try casting it to normal String and it will crash if nil:

var implicitOpt: String!
println("VAR: \(implicitOpt as String)")

The following lines are equal:

var implicitOpt: ImplicitlyUnwrappedOptional<String>
var implicitOpt: String!

ImplicitlyUnwrappedOptional conforms to Printable protocol. So the string interpolation

"\(implicitOpt)"

will call description on ImplicitlyUnwrappedOptional instance and not String instance. Thats why optional is not unwrapped in this case.

0
votes

When you define it as a implicitly unwrapped optional ,the compiler expects it to be initialised and already holding a value .

If you try to do any operation on the variable it will throw you a run time error . You can observe that evaluation doesn't throw any runtime error ,it will give you nil (look at the if loop where it is evaluated)

var implicitOpt:String!
print("VAR: \(implicitOpt)")


var someValue = 10;

if let x = implicitOpt
{
    someValue=1;
}
else
{
    someValue=2
}


var b = implicitOpt;

b = b + "asdf"

let a = implicitOpt + "asdf"