Swift: Unable to make UIApplication conform to my protocol

Question

I'm trying to make UIApplication conform to the following protocol:

protocol URLOpenerProtocol {
    func open(_ url: URL, completionHandler: ((Bool) -> Void)?)
}

I've added the following extension:

extension UIApplication: URLOpenerProtocol {}

However, I am getting a conforming error.

Given the UIApplication already has the following method on it:

UIApplication.shared.open(_ url: URL, completionHandler: ((Bool) -> Void)?)

why won't this conform? The two seem to be indentical to me...

Error is as follows:

Type UIApplication does not conform to protocol URLOpenerProtocol

Answer 1

The actual open method in UIApplication is declared like this with 3 parameters, and the second and third ones happen to be optional (source):

func open(
    _ url: URL,
    options: [UIApplication.OpenExternalURLOptionsKey : Any] = [:],
    completionHandler completion: ((Bool) -> Void)? = nil
)

UIApplication has no two-parameter method with this signature:

func open(_ url: URL, completionHandler: ((Bool) -> Void)?)

It only has the three-parameter one. Therefore, it does not conform to your protocol.

You should add the options parameter into your protocol too, so that UIApplication can conform to it, and I would suggest that you add a protocol extension with the two-parameter version:

protocol URLOpenerProtocol {
    func open(_ url: URL,
              options: [UIApplication.OpenExternalURLOptionsKey : Any],
              completionHandler: ((Bool) -> Void)?)
}

extension URLOpenerProtocol {
    func open(_ url: URL,
              completionHandler: ((Bool) -> Void)?) {
        open(url, options: [:], completionHandler: completionHandler)
    }
}