Abstract parent class calls derived class method

Omer Segal :

So I was wondering about how classes instances are built.

public abstract class test {

    int count;

    public test(){
        count();
        count();
        System.out.println("test" + this.count);
    }
    abstract void count();
}

public class derive extends test{

    int count;

    public derive(){
        System.out.println("derive");
    }
    @Override
    public void count(){
        count++;
    }
}
    public static void main(String[] args) {
    derived o = new derived();
}

The output is:

test0

derive

How come count = 0? and not 2?

rgettman :

The superclass constructor is called first. It calls count() twice. With polymorphism, count() in derive is called, incrementing count to 2. What is incremented is the count variable in derive, because that what the simple name count means in the subclass. The count variable in test is hidden by the count in derive.

However, the print statement refers to the count in scope in the superclass, which is still 0.

Note that when the superclass constructor finishes, then the subclass constructor body can finally execute. This includes giving all instance variables initial values. Here, even though count is already 2, it is "initialized" to 0 anyway. So even if you add a print statement in the subclass constructor, you'll still get 0 for count there too.

To get a count of 2, remove the count in derive and change the count in test to be protected or package-private (no access modifier). This will make count() increment the variable count in test.