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.
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.
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.
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.
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.
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.
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.