Treating a forced downcast as optional will never produce 'nil'

I've been playing around with Swift and discovered that when down casting an object to be inserted into a dictionary, I get a weird warning: Treating a forced downcast to 'String' as optional will never produce 'nil'. If I replace as with as? then the warning goes away.

Treating a forced downcast to 'String' as optional will never produce 'nil'

as

as?

func test() -> AnyObject!
return "Hi!"


var dict = Dictionary<String,String>()
dict["test"]=test() as String


Apple's documentation says the following

Because downcasting can fail, the type cast operator comes in two different forms. The optional form, as?, returns an optional value of the type you are trying to downcast to. The forced form, as, attempts the downcast and force-unwraps the result as a single compound action.

as?

as

I'm unclear as to why using as? instead of as is correct here. Some testing reveals that if I change test() to return an Int instead of a String, the code will quit with an error if I continue using as. If I switch to using as? then the code will continue execution normally and skip that statement (dict will remain empty). However, I'm not sure why this is preferable. In my opinion, I would rather the program quit with an error and let me know that the cast was unsuccessful then simply ignore the erroneous statement and keep executing.

as?

as

as

as?

According to the documentation, I should use the forced form "only when you are sure that the downcast will always succeed." In this case I am sure that the downcast will always succeed since I know test() can only return a String so I would assume this is a perfect situation for the forced form of down casting. So why is the compiler giving me a warning?

test()

import Foundation

String

AnyObject

String

NSString

String

as

as

as!

as!

as

2 Answers
2

Let's take a closer look at your last line, and explode it to see what's happening:

let temporaryAnyObject = test()
let temporaryString = temporaryAnyObject as String
dict["test"] = temporaryString


The error is on the second line, where you are telling the compiler to enforce that temporaryAnyObject is definitely a String. The code will still compile (assuming you don't treat warnings as errors), but will crash if temporaryAnyObject is not actually a String.

temporaryAnyObject

String

temporaryAnyObject

String

The way as works, with no ?, is basically saying "from now on, treat the result of this expression as that type IF the result is actually of that type, otherwise we've become inconsistent and can no longer run.

as

?

The way as? works (with the ?) is saying "from now on, treat the result of this expression as that type IF the result is actually of that type, otherwise the result of this expression is nil.

as?

?

nil

So in my exploded example above, if test() does return a String, then the as downcast succeeds, and temporaryString is now a String. If test() doesn't return a String, but say an Int or anything else not subclassed from String, then the as fails and the code can no longer continue to run.

test()

String

as

temporaryString

String

test()

String

Int

as

This is because, as the developer in complete control, you told the system to behave this way by not putting the optional ? indicator. The as command specifically means that you do not tolerate optional behavior and you require that downcast to work.

?

as

If you had put the ?, then temporaryString would be nil, and the third line would simple remove the "test" key/value pair from the dictionary.

?

temporaryString

nil

This might seem strange, but that's only because this is the opposite default behavior of many languages, like Obj-C, which treat everything as optional by default, and rely on you to place your own checks and asserts.

Edit - Swift 2 Update

Since Swift 2, the forced, failable downcast operator as has been removed, and is replaced with as!, which is much Swiftier. The behavior is the same.

as

as!

let temporaryAnyObject: AnyObject = test(); let temporaryString: String? = temporaryAnyObject as String; dict["test"] = temporaryString

dict["test"] = *

_{You can solve this warning from two angles. 1. The value you return 2. The type that you are expected to return. The other answer speaks about the 1st angle. I'm speaking about the 2nd angle}

This is because you are returning a forced unwrapped and casted value for an optional. The compiler is like, "If you really want to just force cast all optionals then why not just make the expected return parameter to be a non-optional"

For example if you wrote

func returnSomething<T> -> T? // I'm an optional, I can handle nils SAFELY and won't crash.

return UIViewController as! T // will never return a safe nil, will just CRASH




Basically you told yourself (and the compiler) I want to handle nils safely but then in the very next line, you said nah, I don't!!!

nil

The compiler would give a warning:

Treating a forced downcast to 'T' as optional will never produce 'nil'

An alternative is to remove the ?

?

func returnSomething<T>() -> T // I can't handle nils

return UIViewController() as! T // I will crash on nils



Having that said, likely the best way is to not use force cast and just do:

func returnSomething<T>() -> T? // I can handle nils

return UIViewController() as? T // I won't crash on nils



as?

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