How to create an instance of anonymous interface in Kotlin?

余生颓废 提交于 2019-11-27 11:45:27

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")  }
}

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

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

The simplest answer probably is the Kotlin's lambda:

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

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