Swift: Unable to make UIApplication conform to my protocol - swift

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)
}
}