If I have an interface, is there any easy way I can declare a function to make it a public member, but non-overridable? Meaning, it would be exclusively callable and could not be set or overridden by its descendants
interface IFoo {
fun ExecuteOnly(){
// Do Something
}
}
I read a book recently by CommonsWare where this situation was described. and I quote it from there:
"... As a result, anything in an interface hierarchy is permanently open , until you start implementing the interfaces in classes. If that is a problem — if you have some function that you really want to mark as final — use abstract classes, not interfaces..."
You can define an extension function on the interface.
fun IFoo.executeOnly() {
}
It will still be possible for someone to define a member function with that name in a class implementing IFoo but the intention is quite clear. And anyway when using an object via a IFoo reference the IFoo extension will be chosen.
No, you cannot. That's not how Kotlin's interface is implemented.
You can use an abstract class instead
abstract class Foo {
fun executeOnly(){
// Do Something
}
}
Ofcourse You Can... Actually there is not much difference bw kotlin interfaces and abstract classes... simply add a body and a private modifier..
interface MyInterface {
fun triggerTakeMe(){
takeMe()
}
private fun takeMe(){
println("Taken")
}
}
class MyClass : MyInterface
fun main() {
val obj = MyClass()
obj.triggerTakeMe()
}