使用 Akka java API 时 Kotlin 类型推断编译错误

我想在 Kotlin 程序中使用 Akka java API。当我想设置时onCompleteakka 的回调Future，我遇到了 Kotlin 编译器错误，而 java 等效项工作得很好：

val future: Future<Any> = ask(sender, MyActor.Greeting("Saeed"), 5000)

future.onComplete(object : OnComplete<Object>() {
    override fun onComplete(failure: Throwable?, success: Object?) {
        throw UnsupportedOperationException()
    }
}, context.dispatcher())

java代码：

Future<Object> future = ask(sender(), new MyActor.Greeting("Saeed"), 5000);

future.onComplete(new OnComplete<Object>() {
    public void onComplete(Throwable failure, Object result) {
        if (failure != null) {
            System.out.println("We got a failure, handle it here");
        } else {
            System.out.println("result = "+(String) result);
        }
    }
},context().dispatcher());

Kotlin 编译器错误：

Error:(47, 24) Kotlin: Type inference failed: 
fun <U : kotlin.Any!> onComplete(p0: scala.Function1<scala.util.Try<kotlin.Any!>!, U!>!, p1: scala.concurrent.ExecutionContext!): 
kotlin.Unit cannot be applied to (<no name provided>,scala.concurrent.ExecutionContextExecutor!)
Error:(47, 35) Kotlin: Type mismatch: inferred type is <no name provided> but scala.Function1<scala.util.Try<kotlin.Any!>!, scala.runtime.BoxedUnit!>! was expected

我把项目推到了github https://github.com/zarinfam/akka-sample-kotlin.

嗯，错误消息可能有点不清楚，因为有很多 Scala 内容，而且<no name provided>，但它清楚地定义了错误点：您的函数应该接受Any，不是一个Object。以下代码编译没有任何问题：

val future: Future<Any> = ask(sender, MyActor.Greeting("Saeed"), 5000)

future.onComplete(object : OnComplete<Any?>() {
    override fun onComplete(failure: Throwable?, success: Any?) {
        throw UnsupportedOperationException()
    }
}, context.dispatcher())