Swift extend protocol by

Question

Hello so I want to implement something similar in swift

Kotlin example:

class SessionManager(
    private val freeManager: FreeManager
) : FreeManager by freeManager 
{
}

up here FreeManager is Interface I pass reference through constructor and now I can write in extend part FreeManager by freeManager which doesn't require me to implement all methods from interface how can I achieve something similar in swift with protocols?

Can I do something like this:

class SessionManager : FreeManager {

init(freeManager: FreeManager) {
    // assign freeManager to extended protocol instead of implementing 
    //all needed methods from protocol
}
}

Answer 1

You can achieve what you're looking for using a protocol and a protocol extension .

1. Create a protocol name FreeManager with 2 methods,

protocol FreeManager {
    func method1()
    func method2()
}

For now, both method1() and method2() are mandatory to implement by the conforming type .

2. Create a protocol extension and implement the method of protocol that you want to make optional , ie

extension FreeManager {
    func method2() {
        print("This is method2()")
    }
}

In the above code, I've implemented method2() in protocol extension . So, now implementing this method is optional for the conforming type . method1() is still mandatory to implement.

3. Conform class SessionManager to FreeManager

class SessionManager: FreeManager {
    func method1() {
        print("This is method1()")
    }
}

In the above code, I've implemented only method1() .

Answer 2

In Swift you can extend the protocol itself:

protocol Foo {
    func bar()
}

extension Foo {
    func bar() {
        print("bar")
    }
}

class Bar: Foo {
}

let bar = Bar()
bar.bar()