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Home/ Questions/Q 774811
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T19:11:29+00:00 2026-05-14T19:11:29+00:00

I have this simple interface/class: public abstract class Message {} public class Message1 extends

  • 0

I have this simple interface/class:

public abstract class Message {}

public class Message1 extends Message {}

public class Message2 extends Message {}

And an utility class:

public class Utility {
    public void handler(Message m) {
        System.out.println("Interface: Message");
    }

    public void handler(Message1 m) {
        System.out.println("Class: Message1");
    }

    public void handler(Message2 m) {
        System.out.println("Class: Message2");
    }
}

Now, the main class:

public static void main(String[] args) {

    Utility p = new Utility();

    Message1 m1 = new Message1();
    p.handler(m1);

    Message m = (Message) m1;
    p.handler(m);

}

The output is

> Class: Message1
> Interface: Message

I would that p.handler(m) call the method p.handler(m:Message1)

I don’t want use the “manual” command instanceof because I have many cases:

if(m instance of Message1)
p.handler((Message1)m)
else if (m instanceof Message2)
p.handler((Message2)m)
...

If I call m.getClass() I obtain “mypackage.Message1”, so the subclass and not the superclass.

I try with this code (use reflection):

p.handler(m.getClass().cast(m));

But the output is

> Interface: Message

So, this is my problem. I would do a runtime cast of superclass object to subclassobject without use the “code command” istanceof.

I would a right command like this:

p.handler((m.getclass)m);

How can I obtain it? It’s possible?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T19:11:30+00:00Added an answer on May 14, 2026 at 7:11 pm

    Java will call the method on the basis of information known at compile time. What you could do is add a method to the interface that calls the correct handler method for the object.

    public abstract class Message {
    
        public abstract void callHandler(Utility utility);
    
    }
    
    public class Message1 extends Message{
    
        public void callHandler(Utility utility) {
            utility.handler(this);
        }
    }
    

    Your calls to the handler become:

    Message m=(Message) m1;
    m.callHandler(p);
    

    which now calls Utility::handler(Message1) even though the reference in main is of type of the Message interface.

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