从Java代码在Scala Future中包装String

Question

I have the following method for testing:

public scala.concurrent.Future<String> send() { 
  return "OK"; //error - should wrap with future
} 

How can I make it return a scala future (this is java code, so didn't find any example)?

Answer 1

It's easy if you are working with the Scala 2.12.x library.

In Scala you would write Future(doSomeStuff()) . Or simply Future.successful("OK") if you really only want to wrap a literal in a Future . In Java that translates to one of these:

import scala.concurrent.*;

public Future<String> send1() { 
    return Future.apply( () -> doSomeStuff(), ExecutionContext.global());
}

public Future<String> send2(ExecutionContext ec) { 
    return Future.apply( () -> doSomeStuff(), ec);
}

public Future<String> send3() { 
    return Future.successful("OK");
}

If you have more complex code to run than simply "OK" you'll have to use send1 or send2 . Which of those you want depends on whether you want to use the default ExecutionContext or let the caller decide.

---

If your Scala version is lower than 2.12.0, the send1 method from above will look something like this:

import scala.concurrent.*;
import scala.runtime.AbstractFunction0;

public Future<String> send1() { 
    return Future$.MODULE$.apply( new AbstractFunction0<String>() {
        public String apply() {
            return doSomeStuff();
        }
    }, ExecutionContext$.MODULE$.global());
}