Java: A synchronized method in the superclass acquires the same lock as one in the subclass, right?

Question

class A {
    public synchronized void myOneMethod() {
        // ...
    }
}

class B extends A {
    public synchronized void myOtherMethod() {
        // ...
    }
}
         


        

Answer 1

If you want to be more explicit about your locking, you could do something like this:

class A {

    protected final Object  mutex = new Object();
    public void myOneMethod() {
        synchronized (mutex) {
            // ...
        }
    }
}

class B extends A {
    public  void myOtherMethod() {
        synchronized (mutex) {
            // ...
        }
    }
}

In fact, this pattern is recommended by Brian Goetz in Java Concurrency in Practice, section 4.2.1 "The Java monitor pattern". That way you know exactly where your monitor is coming from.