If two threads are waiting to enter a synchronized method, when the mutex is released do they execute in the order they arrived?

Question

If I have a synchronized method and two threads are waiting to enter it they seem to enter the thread Last In First Executed. Is there a way to make this First In First Executed?

This is the unit test that I'm using:

package com.test.thread;

import org.apache.log4j.Logger;
import org.junit.Test;

public class ThreadTest {
    private static final Logger log = Logger.getLogger(ThreadTest.class);

    @Test
    public void testThreading() throws InterruptedException {
        Thread t1 = new Thread(new Runnable() {    
            public void run() { synchd("1"); }
        });
        Thread t2 = new Thread(new Runnable() {    
            public void run() { synchd("2"); }
        });
        Thread t3 = new Thread(new Runnable() {    
            public void run() { synchd("3"); }
        });

        t3.start();
        Thread.sleep(5);
        t1.start();
        t2.start();

        Thread.sleep(12000);
    }

    public static synchronized void synchd(String output) {
        log.debug(output);
        try {
            Thread.sleep(3000);
        } catch (InterruptedException e) {
            // do nothing
        }

    }
}

The output for this is always 3, 2, 1, and I'd like to find a way for it to be 3, 1, 2.

Answer 1

No, they are not executed in their order of arrival.

The execution order depends on lots of parameters relative to threads scheduling and is most often unpredictable.