In Kotlin's interfaces with default implementations, there's no way to constrain the interface to a ...

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[iex]

In Kotlin's interfaces with default implementations, there's no way to constrain the interface to a specific implementation / access it, right?

Edit: Here's a snippet in Swift of what I mean. I'm almost sure it's not possible in Kotlin, but just want to confirm (and see what alternative can be used)

protocol ChildViewController where Self: UIViewController { // <--ChildViewController is "interface" and it's constrained to a specific class

    func foo()
}

extension ChildViewController {

    func foo() { <-- this is basically default interface implementation
        print(self.view) <-- here we access a field of the concrete type, i.e. UIViewController
    }

}

[thana]

No, maybe sealed classes can help you, or maybe use an abstract class with a final implementation of some method (im unsure what you mean)

[iex]

no that's not it. I added an example in Swift of what I mean with some explanatory comments.

[elizarov]

One way to do it in Kotlin is:

interface ChildViewController<Self: UIViewController> {
    val self: Self
}

fun <Self: UIViewController> ChildViewController<Self>.foo() {
    println(self.view)
}

[iex]

@elizarov thanks! I rewrote it using more familiar Android things:

interface MyInterface<T: Fragment> {
    val self: T
}
fun <T: Fragment> MyInterface<T>.foo() {
    println(self.view)
}

But I still need to add a method to the class (Fragment) in this case to support this:

override val self: RegisterFragment get() = this

Which is not as nice...