如何在 Kotlin 中创建匿名接口的实例

发布于06月07日

我有一个第三方Java库，其中包含一个具有如下接口的对象:

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"

推荐答案

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")  }
}