Is it possible to satisfy Swift protocol and add defaulted arguments?

Question

If you have protocol like so:

protocol Messaging {
    func sendMessage(message: String)
}

Is there any way to satisfy it in a class like so:

class Messager: Messaging {
    func sendMessage(message: String, count: Int = 1) {}
}

This would be nice to have, as the resulting signature of the protocol is satisfied by adding the defaulted parameter. Is there any way to get this to work with Swift 2?

This is a simplified example. Let's say, for the sake of argument, that the protocol is fixed. A solution can only update the Messager class. My goal is to be able to call sendMessage() like so:

let m: Messaging = Messager()
m.sendMessage("")

The only way I found to accomplish this (and satisfy the compiler) is with overloading like so:

class Messager: Messaging {
    func sendMessage(message: String) {
        self.sendMessage(message, count: 1)
    }

    func sendMessage(message: String, count: Int = 1) {}
}

The problem with this approach is that my defaults are then specified in two places and I lose the main advantage of Swift's default parameters.

Answer 1

in Swift 3 you could use extensions to solve that, however its a bit ugly. Hope for a better solution in next swift versions.

import UIKit

protocol TestProtocol {
    func testFunction(a:Int, b:Int?) -> String
}

extension TestProtocol
{
    func testFunction(a:Int, b:Int? = nil) -> String {
        return testFunction(a:a, b:b)
    }
}

class TestClass: TestProtocol
{
    func testFunction(a:Int, b:Int?) -> String {
        return "a:\(a), b:\(b)"
    }
}

func testit(testProtocol: TestProtocol) {
    print(testProtocol.testFunction(a:10)) // will print a:10, b:nil
    print(testProtocol.testFunction(a:10, b:20)) // will print a:10, b:Optional(20)
}

let t = TestClass()
testit(testProtocol: t)