instansof operator.

[Jan Osykowski]

Hey,

I have read that instanceof operator can not be used on the instances/classes that belong to different inheritance tree and that it will result in compile error. But now I have come across this example:

 System.out.println(new HashMap() instanceof Collection);   

And as I look in the book HashMap and Collection are in completly different inheritance tree but the above code still works. Can anyone explain why?

Cheers,
Jan.

[Ireneusz Kordal]

Hi,

Collection is an interface, not a class.
Interfaces are different from classes - they are not in the class inheritance tree, there are 'oudside' of this tree.
Where an interface is on the right side of instanceof operator, the operator tests if given object implements that interface.
Virtually all classes can implement any interface, and compiler cannot reject this test at compile time.


Look at this example:

class HashMappp{}

interface Collectionnn {}

class HashMappp_Collection extends HashMappp implements Collectionnn{
}

class OtherClass implements Collectionnn{
}

public class Main {
    public static void main(String[] args) {
        Object o1 = new HashMappp_Collection();
        Object o2 = new HashMappp();
        Object o3 = new OtherClass();

        System.out.println( o1  instanceof Collectionnn );
        System.out.println( o2 instanceof Collectionnn );
        System.out.println( o3 instanceof Collectionnn );
    }
}


true
false
true

Here we have:
1. interface Collectionnn
2. two clasess: HashMappp, HashMappp_Collection in one 'inheritance subtree'
3. third class: OtherClass that is in the other 'inheritance subtree' than classes from point 2.
HashMappp doesn't implement Collectionnn, but two other classes do.

[Jan Osykowski]

Thanks you for the answer and a great example! Yeah, that's straightforward now. A class can implement any interface so instanceof operator can not reject it !

[Avishkar Nikale]

I think better explanation is that a subclass of the given class can implement the given interface hence the compiler creates no issues.

If you make the class final & there is no implementation of the interface in the hierarchy it will throw an compilation error.

E.g 1 will compile but Eg. 2 wont

1.

import java.io.Serializable;

class One {}

  
public class Main {  
    public static void main(String[] args) {  
    	One o1 = new One();  
      
  
        System.out.println( o1  instanceof Serializable );  
         
    }  
} 

2.

import java.io.Serializable;

final class One {}

  
public class Main {  
    public static void main(String[] args) {  
    	One o1 = new One();  
      
  
        System.out.println( o1  instanceof Serializable );  
         
    }  
}  

[Dieter Quickfend]

Good find Avishkar! Thanks for the clarification.

[Avishkar Nikale]

Dieter Quickfend wrote:Good find Avishkar! Thanks for the clarification.

Thanks Dieter,

All the best with your exam.