符合通用函数中使用的协议和类的类型的变量

Question

I want to declare a variable

var specialVC: UIViewController & MyProtocol.

And I have a function

func doStuff<T: UIViewController & MyProtocol> { ... }

However, when I try to pass in my variable into the doStuff, it says UIViewController doesn't conform to MyProtocol.

class MyClass: UIViewController {

    override func viewDidLoad() {
      super.viewDidLoad()
      var specialVC: UIViewController & MyProtocol
      doStuff(specialVC)
    }

    func doStuff<T: UIViewController & MyProtocol>(_ vc: T) {}

}

error: Argument type 'UIViewController' does not conform to expected type 'MyProtocol'

--- Update ---

After looking at Protocol doesn't conform to itself? , I am able to create an extension that specifies a class that conforms to a protocol. However, I won't be able to call doStuff() from this extension.

internal extension MyProtocol where Self: UIViewController {
     // call doStuff(self) here somehow?
}

Answer 1

There is nothing about your function that needs to be a generic. Just use the normal types-and-supertypes mechanism (polymorphism, inheritance, whatever you like to call it). Just type your parameter as the supertype; this tells the compiler that it inherits all features of the supertype.

protocol MyProtocol : UIViewController { // Swift 5 syntax
    var thingy : String { get set }
}
class MyViewController : UIViewController, MyProtocol {
    var thingy = "howdy"
}
func doStuff(_ vc: MyProtocol) {
    print(vc.title) // legal, because we know it's a view controller
    print(vc.thingy) // legal, because we know it's a MyProtocol
}