Question

I have one static Semaphore instance.

    Semaphore semaphore = new Semaphore(1);

Now I have two Threads (Sending Thread and Receiving Thread)

Sending Thread:

public class SendingThread implements Runnable {

    private Semaphore semaphore;

    public SendingThread(Semaphore semaphore){
        this.semaphore = semaphore;
    }

    @Override
    public void run() {
        try {
            System.out.println("1");
            Thread.sleep(4000);
            this.semaphore.acquire();

        } catch (InterruptedException e) {
            Thread.currentThread().interrupt();
            e.printStackTrace();
        }   
    }
}

and receiving Thread:

public class RecievingThread implements Runnable {

    private Semaphore semaphore;

    public RecievingThread(Semaphore semaphore){
        this.semaphore = semaphore;
    }
    @Override
    public void run() {
        System.out.println("2");
         this.semaphore.release();
            System.out.println("3");
    }
} 

When I start this 2 threads according to my understanding Receiving Thread will wait for 4 second till
Sending Thread will notify it that Receiving Thread can continue. It means that System.out.println("3"); will be printed with 4 second delay, but when I run this code all three values are printed immediately. Why?

Am i missing something?

Answer 1

A new Semaphore(1) has 1 initial permit and thus allows for one immediate acquire to go through. Furthermore since a release is always allowed both threads proceeds immediately.

To force one thing to happen before the other, you can use new Semaphore(0). This will force the thread calling acquire to wait for the thread executing release.