How to create an instance of anonymous interface in Kotlin?

问题

I have a third party Java library which an object with interface like this:

public interface Handler<C> {
  void call(C context) throws Exception;
}

How can I concisely implement it in Kotlin similar to Java anonymous class like this:

Handler<MyContext> handler = new Handler<MyContext> {
   @Override
   public void call(MyContext context) throws Exception {
      System.out.println("Hello world");
   }
}

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

回答1:

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

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

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")  }
}

回答2:

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() {}

                        override fun OnPermissionsDeniedForever() {}
})

回答3:

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

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

回答4:

The simplest answer probably is the Kotlin's lambda:

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

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

来源：https://stackoverflow.com/questions/37672023/how-to-create-an-instance-of-anonymous-interface-in-kotlin

标签

java

lambda

kotlin

anonymous