How to create an instance of anonymous interface in Kotlin?

Question

Assuming the interface has only a single method you can make use of SAM.

val handler = Handler<String> { println("Hello: $it") }

Since version 1.4 Kotlin supports SAM for interfaces defined in Kotlin. That requires prefixing the interface keyword with fun

fun interface Handler<C> {
  fun call(context: C);
}

If you have a method that accepts a handler then you can even omit type arguments:

fun acceptHandler(handler:Handler<String>){}

acceptHandler(Handler { println("Hello: $it") })

acceptHandler({ println("Hello: $it") })

acceptHandler { println("Hello: $it") }

If the interface has more than one method the syntax is a bit more verbose:

val handler = object: Handler2<String> {
    override fun call(context: String?) { println("Call: $context") }
    override fun run(context: String?) { println("Run: $context")  }
}

I had a case where I did not want to create a var for it but do it inline. The way I achieved it is

funA(object: InterfaceListener {
                        override fun OnMethod1() {}

                        override fun OnMethod2() {}
})

     val obj = object : MyInterface {
         override fun function1(arg:Int) { ... }

         override fun function12(arg:Int,arg:Int) { ... }
     }

As of Kotlin 1.4 you can declare a functional interface:

fun interface Handler<C> {
  fun call(context: C);
}

and then you can create one concisely:

val handler = Handler<String> {
  println("Handling $it")
}

Demo

The simplest answer probably is the Kotlin's lambda:

val handler = Handler<MyContext> {
  println("Hello world")
}

handler.call(myContext) // Prints "Hello world"