Method from constructor

[Andry Dub]

Hi! Why my code

public class Derived extends Base {
    int i = -1;
 
    public static void main(String argv[]) {
        Base b = new Derived();
        System.out.println(b.i);
        b.amethod();
    }
 
    public void amethod() {
        System.out.println("Derived.amethod()");
    }
}
 
class Base {
    int i = 99;
 
    public void amethod() {
        System.out.println("Base.amethod()");
    }
 
    Base() {
        amethod();
    }
}

gives output

Derived.amethod() --- why? "Base.amethod()" expected
99
Derived.amethod()

Base constructor can't know anything about it's children.

[Kelvin Chenhao Lim]

Originally posted by Andry Dub:
Base constructor can't know anything about it's children.

You're right, it doesn't. But thanks to the magic of polymorphism, the Base constructor's call to amethod() will be bound at runtime to the overridden implementation in Derived. In fact, this doesn't only apply to constructors; this is true of any call by a Base method to another Base method, unless the latter cannot be overridden (e.g. if it's final or private).

Among other things, this mechanism allows the use of the template design pattern in Java.

[Andry Dub]

Thank you, Kelvin! It is very valuable information for me.