6.6.5 Example: Default-Access Fields, Methods, and Constructors

If none of the access modifiers public, protected, or private are specified, a class member or constructor is accessible throughout the package that contains the declaration of the class in which the class member is declared, but the class member or constructor is not accessible in any other package. If a public class has a method or constructor with default access, then this method or constructor is not accessible to or inherited by a subclass declared outside this package.

For example, if we have:

package points;

public class Point {
	public int x, y;
	void move(int dx, int dy) { x += dx; y += dy; }
	public void moveAlso(int dx, int dy) { move(dx, dy); }
}

then a subclass in another package may declare an unrelated move method, with the same signature (§8.4.2) and return type, but because the original move method is not accessible from package morepoints super may not be used:

package morepoints;

public class PlusPoint extends points.Point {
	public void move(int dx, int dy) {
		super.move(dx, dy);								// compile-time error
		moveAlso(dx, dy);
	}
}

Because move of Point is not overridden by move in PlusPoint, the method moveAlso in Point never calls the method move in PlusPoint.

Thus if you delete the super.move call from PlusPoint and execute the test program:

import points.Point;

import morepoints.PlusPoint;

class Test {

    public static void main(String[] args) {
        PlusPoint pp = new PlusPoint();
        pp.move(1, 1);
    }

}

it terminates normally. If move of Point were overridden by move in PlusPoint, then this program would recurse infinitely, until a StackoverflowError occurred.