I\'d like to return a UIViewController conforming to MyProtocol from a method, so I\'m using the method signature:
func myMethod<
func myMethod() -> T
This declaration says: There exists a function called myMethod, such that myMethod returns some specific T where T is a subtype of UIViewController and also MyProtocol. This does not say what type T actually is, and it does not say that there is only one such myMethod. There can be many if there are many type that are both subclasses of UIViewController and conform to MyProtocol. Every one of those types creates a new version of myMethod (really a new solution to the assertion myMethod makes, that such a function does exist).
This is not the same thing as:
func myMethod() -> UIViewController
That says: The function myMethod returns any subtype of UIViewController.
There is no way in Swift to express "any type that is a subclass of UIViewController and is a subtype of MyProtocol." You can only discuss a specific type that meets that criterial. Swift can't combine classes and protocols this way; it's just a current limitation of the language, not a deep design issue.
The specific versus any is the issue. There are many functions that satisfy your myMethod declaration. Every T you can plug in that conforms to the rules would be a candidate. So when you say myMethod(), the compiler doesn't know which specific T you mean.
(I was going to expand this answer to provide it in less type-theory, more "how do you do it in code" terms, but donnywals already has an excellent version of that.)
* To your edited question *
func myMethod() -> T where T : UIViewController, T : MyProtocol {
return MyViewController() as! T // why is the cast necessary?
}
T is a specific type decided by the caller. It is not "any type that conforms" it is "some specific, concrete type that conforms." Consider the case that you called:
let vc: SomeOtherViewController = myMethod()
In this case, T is SomeOtherViewController. MyViewController is not that type, so what you're doing with the as! cast is dangerous.