Using Lambda in Thread

Question

ThreadWorker does not execute when I use method reference for creating new object in the constructor of Thread or passing Lambda for creation of new Object.

But it works fine when I create ThreadWorker object separately and I pass that to Thread class.

public class RunnableImpl {
    public static void main(String[] args) throws InterruptedException {
        ThreadWorker th= new ThreadWorker();
        Thread t1 = new Thread(th);
        t1.start();
        t1.join();
        System.out.println("Main method terminated");
    }
}

class ThreadWorker implements Runnable {

    @Override
    public void run() {
        int[] arr = { 1, 4, 8, 9, 1, 0, 4, 5, 4 };
        System.out.println(Arrays.stream(arr).sum());
    }
}

If I use lambda for example:

Thread t1 = new Thread(ThreadWorker :: new);

or

Thread t1 = new Thread(() ->new ThreadWorker());

then there is no output but If I create ThreadWorker Object separately then the program is working fine.

Can someone please let me know, how it is possible?

Thread expects a runnable Object, not a lambda to create a runnable Object. Use new ThreadWorker() insted of () ->new ThreadWorker() or ThreadWorker :: new. — dan1st
– dan1st, Commented Jun 3, 2019 at 7:18

Eran · Accepted Answer · 2019-06-03 07:23:06Z

6

Both

new Thread(() -> new ThreadWorker());

and

new Thread(ThreadWorker::new);

create a Thread whose Runnable instance's run() method simply creates a ThreadWorker instance, and does nothing with it. The run() method of ThreadWorker is not executed.

They are equivalent to passing the following anonymous class instance:

Thread t = new Thread(new Runnable() {
                                         public void run() {
                                             new ThreadWorker ();
                                         }
                                     });

If you want to use a lambda expression, you need:

new Thread(() -> new ThreadWorker().run());

If you want a method reference, you need:

new Thread(new ThreadWorker()::run);

That said, the following is much simpler:

new Thread(new ThreadWorker());

edited Jun 3, 2019 at 7:23

answered Jun 3, 2019 at 7:12

Eran

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6 Comments

Antoniossss Over a year ago

Didnt know about last variant thus +1 (which seams kind of obvious as you can reference instance methods no problem)

Atishay Jain Over a year ago

But in this case,ThreadWorker th= new ThreadWorker();Thread t1 = new Thread(th); we are doing the same thing, creating Object and then passing to Thread class.

Eran Over a year ago

@AtishayJain In the case of Thread t1 = new Thread(th);, you are passing an instance of ThreadWorker, which implements Runnable, so the thread will execute its run() method when started.

Eran Over a year ago

@AtishayJain No, () -> new ThreadWorker() is not the same as new ThreadWorker(). One is an instance of ThreadWorker, which implements Runnable. The other is a lambda expression implementing a functional interface whose method's body creates a ThreadWorker instance. That lambda expression happens to fit the Runnable interface, which allows you to pass it to the Thread constructor, but it behaves differently.

Atishay Jain Over a year ago

@Eran I got it now, So in short Lambda is a method body, like the action which can be performed and it will be performed by a functional interface method which is run() method in this case. Thanks a ton for your quick help.

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