Is there a Java equivalent to C#'s 'yield' keyword?

JavaYieldYield Return

Java Problem Overview


I know there is no direct equivalent in Java itself, but perhaps a third party?

It is really convenient. Currently I'd like to implement an iterator that yields all nodes in a tree, which is about five lines of code with yield.

Java Solutions


Solution 1 - Java

The two options I know of is Aviad Ben Dov's infomancers-collections library from 2007 and Jim Blackler's YieldAdapter library from 2008 (which is also mentioned in the other answer).

Both will allow you to write code with yield return-like construct in Java, so both will satisfy your request. The notable differences between the two are:

Mechanics

Aviad's library is using bytecode manipulation while Jim's uses multithreading. Depending on your needs, each may have its own advantages and disadvantages. It's likely Aviad's solution is faster, while Jim's is more portable (for example, I don't think Aviad's library will work on Android).

Interface

Aviad's library has a cleaner interface - here's an example:

Iterable<Integer> it = new Yielder<Integer>() {
    @Override protected void yieldNextCore() {
        for (int i = 0; i < 10; i++) {
            yieldReturn(i);
            if (i == 5) yieldBreak();
        }
    }
};

While Jim's is way more complicated, requiring you to adept a generic Collector which has a collect(ResultHandler) method... ugh. However, you could use something like this wrapper around Jim's code by Zoom Information which greatly simplifies that:

Iterable<Integer> it = new Generator<Integer>() {
    @Override protected void run() {
        for (int i = 0; i < 10; i++) {
            yield(i);
            if (i == 5) return;
        }
    }
};
License

Aviad's solution is BSD.

Jim's solution is public domain, and so is its wrapper mentioned above.

Solution 2 - Java

Both of these approaches can be made a bit cleaner now Java has Lambdas. You can do something like

public Yielderable<Integer> oneToFive() {
    return yield -> {
        for (int i = 1; i < 10; i++) {
            if (i == 6) yield.breaking();
            yield.returning(i);
        }
    };
}

I explained a bit more here.

Solution 3 - Java

I know it's a very old question here, and there are two ways described above:

  • bytecode manipulation that's not that easy while porting;
  • thread-based yield that obviously has resource costs.

However, there is another, the third and probably the most natural, way of implementing the yield generator in Java that is the closest implementation to what C# 2.0+ compilers do for yield return/break generation: lombok-pg. It's fully based on a state machine, and requires tight cooperation with javac to manipulate the source code AST. Unfortunately, the lombok-pg support seems to be discontinued (no repository activity for more than a year or two), and the original Project Lombok unfortunately lacks the yield feature (it has better IDE like Eclipse, IntelliJ IDEA support, though).

Solution 4 - Java

Stream.iterate(seed, seedOperator).limit(n).foreach(action) is not the same as yield operator, but it may be usefull to write your own generators this way:

import java.util.stream.Stream;
public class Test01 {
    private static void myFoo(int someVar){
        //do some work
        System.out.println(someVar);
    }
    private static void myFoo2(){
        //do some work
        System.out.println("some work");
    }
    public static void main(String[] args) {
        Stream.iterate(1, x -> x + 1).limit(15).forEach(Test01::myFoo);     //var1
        Stream.iterate(1, x -> x + 1).limit(10).forEach(item -> myFoo2());  //var2
    }
}

Solution 5 - Java

I'd also suggest if you're already using RXJava in your project to use an Observable as a "yielder". It can be used in a similar fashion if you make your own Observable.

public class Example extends Observable<String> {

	public static void main(String[] args) {
		new Example().blockingSubscribe(System.out::println); // "a", "b", "c", "d"
	}
    
	@Override
	protected void subscribeActual(Observer<? super String> observer) {
		observer.onNext("a"); // yield
		observer.onNext("b"); // yield
		observer.onNext("c"); // yield
		observer.onNext("d"); // yield
		observer.onComplete(); // finish
	}
}

Observables can be transformed into iterators so you can even use them in more traditional for loops. Also RXJava gives you really powerful tools, but if you only need something simple then maybe this would be an overkill.

Solution 6 - Java

I just published another (MIT-licensed) solution here, which launches the producer in a separate thread, and sets up a bounded queue between the producer and the consumer, allowing for buffering, flow control, and parallel pipelining between producer and consumer (so that the consumer can be working on consuming the previous item while the producer is working on producing the next item).

You can use this anonymous inner class form:

Iterable<T> iterable = new Producer<T>(queueSize) {
    @Override
    public void producer() {
        produce(someT);
    }
};

for example:

for (Integer item : new Producer<Integer>(/* queueSize = */ 5) {
    @Override
    public void producer() {
        for (int i = 0; i < 20; i++) {
            System.out.println("Producing " + i);
            produce(i);
        }
        System.out.println("Producer exiting");
    }
}) {
    System.out.println("  Consuming " + item);
    Thread.sleep(200);
}

Or you can use lambda notation to cut down on boilerplate:

for (Integer item : new Producer<Integer>(/* queueSize = */ 5, producer -> {
    for (int i = 0; i < 20; i++) {
        System.out.println("Producing " + i);
        producer.produce(i);
    }
    System.out.println("Producer exiting");
})) {
    System.out.println("  Consuming " + item);
    Thread.sleep(200);
}

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
Questionripper234View Question on Stackoverflow
Solution 1 - JavaOakView Answer on Stackoverflow
Solution 2 - JavabenjiweberView Answer on Stackoverflow
Solution 3 - JavaLyubomyr ShaydarivView Answer on Stackoverflow
Solution 4 - JavaAndrey LavrukhinView Answer on Stackoverflow
Solution 5 - JavaGyőri SándorView Answer on Stackoverflow
Solution 6 - JavaLuke HutchisonView Answer on Stackoverflow