Pass scala function as java functional interface argument

Question

I have a java constructor which takes a functional interface as a parameter:

public ConsumerService(QueueName queue, Consumer<T> function) {
    super(queue);
    this.function = function;
}

I'm trying to use this constructor in scala but the compiler complains, saying it cannot resolve the constructor. I've tried the following ways:

val consumer = new ConsumerService[String](QueueName.CONSUME, this.process _)
val consumer = new ConsumerService[String](QueueName.PRODUCE, (s: String) => this.process(s))

I've also tried to cast the function at runtime - but that leaves me with a ClassCastException:

val consumer = new ConsumerService[String](QueueName.CONSUME, (this.process _).asInstanceOf[Consumer[String]])

How can I pass a scala function as a java functional interface parameter?

http://stackoverflow.com/questions/29684894/is-there-a-standard-way-to-convert-a-java-util-function-consumert-into-a-java but there is no real answer there either... — Alec, Jul 19 '16 at 18:22

score 13 · Accepted Answer · answered Jul 19 '16 at 18:37

13

You need to create a Consumer:

val consumer = new ConsumerService[String](QueueName.CONSUME, 
  new Consumer[String]() { override def accept(s: String) = process(s) })

answered Jul 19 '16 at 18:37

Jean Logeart

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score 5 · Answer 2 · answered Nov 30 '17 at 13:11

Another way is to use a following library:

https://github.com/scala/scala-java8-compat.

With that in your project once you import:

import scala.compat.java8.FunctionConverters._

you can now write:

asJavaConsumer[String](s => process(s))

which in your code should go like this:

val consumer = new ConsumerService[String]( QueueName.CONSUME, asJavaConsumer[String](s => process(s)) )

Pass scala function as java functional interface argument

2 Answers2