Make property of type and also conform to protocol in Swift

Question

I would like to make a property that is of a certain type and also conforms to a protocol, which I would have done in Objective-C like this:

@property (nonat         


        

Answer 1

I can't think of a good way to express this in Swift. The syntax for a type is:

type → array-type­ | dictionary-type­ | function-type­ | type-identifier­ | tuple-type­ | optional-type­ | implicitly-unwrapped-optional-type­ | protocol-composition-type­ | metatype-type­

What you're looking for would be a kind of protocol-combination-type that also accepts a base class. (Protocol-combination-type is protocol<Proto1, Proto2, Proto3, …>.) However, this is not currently possible.

Protocols with associated type requirements are allowed to have typealiases that specify a required base class and required protocols, but these also require the types to be known at compile-time, so it's unlikely to be a viable option for you.

If you're really into it, you can define a protocol with the parts of the interface of UIViewController that you use, use an extension to add conformance, and then use protocol<UIViewControllerProtocol, CustomProtocol>.

protocol UIViewControllerProtocol {
    func isViewLoaded() -> Bool
    var title: String? { get set }
    // any other interesting property/method
}

extension UIViewController: UIViewControllerProtocol {}

class MyClass {
    var controller: protocol<UIViewControllerProtocol, CustomProtocol>
}

Answer 2

This is now possible using the built-in composition:

var children = [UIViewController & NavigationScrollable]()