CODEWITHSUNDEEP
androidkotlin
Algar
Algar

Reputation: 5984

Kotlin extension on interface

Let's say I have an interface Base and that we implement that interface in class Base1 : Base.

I would expect an extension function in the form

fun ArrayList<Base>.myFun()

to also work on arrayListOf(Base1(), Base1()).myFun(), but it doesn't. It requires the list to be of type Base instead of Base1.

Isn't this really strange? Or am I just missing something?

And, what are my options to write a function available on all subclasses of an interface?

Thanks!

Upvotes: 12

Views: 14311

Answers (4)

Francisco Durdin Garcia
Francisco Durdin Garcia

Reputation: 13327

Lets say that you have:

interface Base
class Base1: Base
class Base2: Base

To have a fun for each class that extends your interface, you should create a class with a generic extension:

fun <T : Base> ArrayList<T>.myFun() = arrayListOf<T>()

In this case what you are doing is do an extension on any ArrayList which container class would be any class that extends from your Base interface. In the same way it will return an arraylist of that type.

In code this will look like:

fun myCode(){
    val array1 : ArrayList<Base1> = arrayListOf(Base1()).myFun()
    val array2 : ArrayList<Base2> = arrayListOf(Base2()).myFun()
}

You can read more about it in the official documentation, it has a lot of generic examples too: https://kotlinlang.org/docs/reference/extensions.html

Upvotes: 3

ruX
ruX

Reputation: 7472

You need to allow extension function to accept child implementation 

interface Base
class Base1: Base

fun ArrayList<out Base>.myFun() = println(toString())

fun main(args: Array<String>) {
    arrayListOf(Base1(), Base1()).myFun()

}

Upvotes: 12

Sweeper
Sweeper

Reputation: 271420

This:

fun ArrayList<Base>.myFun()

is an extension only on ArrayList<Base>.

You can make the function generic to support any implementers of Base:

fun <T : Base> ArrayList<T>.myFunc() {}

Usage:

ArrayList<Base1>().myFunc()

Upvotes: 0

Dmitry
Dmitry

Reputation: 2526

This should work with array list of any type:

fun <T> ArrayList<T>.myFun()

Upvotes: -1

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