Question

Why is the super() constructor recommended to be called when extending any of the RIM Screen classes? It was my understanding that the no-arg constructor of any super class was implicitly called from any class that extends it, am I wrong?

Answer 1

super() will be called by default constructor no matter if you implement it or not.

It’s easy to test. Just a couple classes:

class ClassA {
    public ClassA() {
        UiApplication.getUiApplication().invokeLater(new Runnable() {
            public void run() {
                Dialog.inform("ClassA()");
            }
        });
    }
}

class ClassB extends ClassA {
    public ClassB() {
        super();
        UiApplication.getUiApplication().invokeLater(new Runnable() {
            public void run() {
                Dialog.inform("ClassB()");
            }
        });
    }
}

class ClassC extends ClassA {

}

class ClassD extends ClassA {
    public ClassD() {
        UiApplication.getUiApplication().invokeLater(new Runnable() {
            public void run() {
                Dialog.inform("ClassD()");
            }
        });
    }
}

And test app:

class Scr extends MainScreen {

    protected void makeMenu(Menu menu, int instance) {
        super.makeMenu(menu, instance);

        menu.add(new MenuItem("class A", 0, 0) {
            public void run() {
                ClassA a = new ClassA();
            }
        });
        menu.add(new MenuItem("class B", 0, 0) {
            public void run() {
                ClassB b = new ClassB();
            }
        });
        menu.add(new MenuItem("class C", 0, 0) {
            public void run() {
                ClassC c = new ClassC();
            }
        });
        menu.add(new MenuItem("class D", 0, 0) {
            public void run() {
                ClassD d = new ClassD();
            }
        });
    }
}

So you see, there is no difference between dialog calls in ClassA() and ClassC(), also no diffs between ClassB() and ClassD().