using instanceof operator in Java

Question

I have a base class called class Base and two children classes

class A extends Base

and

class B extends Base

I have a method foo in Base.

Rather than putting the implementation of foo in class A and class B, so that I can do

void foo (Object o)
{
    // A's implementation
    assert o instanceof A;     
}


 void foo (Object o)
 {
     // B's implementation
     assert o instanceof B; 
 }

Is there anyway to put foo in Base, and still still be able to check for the runtime class? I've thought of something like this:

 void foo (Object o)
 {
    // Check that o is instanceof a runtime class
    assert o instanceof this.getClass(); // ????
 }

Thanks.

yes, it's possible with the getClass().isInstance(obj) method, but why would you like to do this? If the foo overload is specific for each subclass, then you should do implement that in the subclass. Making a superclass aware of it's subclasses is bad. If the superclass needs to do something that only the subclass will know, consider using an abstract method declaration instead. — maasg
– maasg, Commented Jun 18, 2012 at 15:27

JB Nizet · Accepted Answer · 2012-06-18 15:18:46Z

6

You can implement your method like this:

public void foo() {
    if (this instanceof A) {
        // implementation for A
    }
    else if (this instanceof B) {
        // implementation for B
    }
}

But the point of polymorphism is to put the A implementation in A, so that this implementation can use A's private fields to implement the method (same for B, or course).

answered Jun 18, 2012 at 15:18

JB Nizet

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Comments

meriton · Accepted Answer · 2012-06-18 15:24:27Z

2

getClass().isInstance(o)

That said, perhaps you want to constrain the type in a way the compiler can check? Generics can do that:

class Base<T extends Base<B>> {
    void foo(T o) { ... }
}

class A extends Base<A> {
    @Override void foo(A o) { ... }
}

Then,

new B().foo(new A());

will not compile.

edited Jun 18, 2012 at 15:24

answered Jun 18, 2012 at 15:17

meriton

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Comments

anstarovoyt · Accepted Answer · 2012-06-18 15:27:44Z

1

There is class method isAssignableFrom()

   getClass().isAssignableFrom(o.getClass())

edited Jun 18, 2012 at 15:27

answered Jun 18, 2012 at 15:16

anstarovoyt

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3 Comments

meriton Over a year ago

Erm, shouldn't it be the other way around? As in getClass().isAssignableFrom(o.getClass())?

user85421 Over a year ago

should it not be the reverse?

anstarovoyt Over a year ago

Yes, it is my mistake. Updated

user85421 · Accepted Answer · 2012-06-18 15:39:04Z

1

instanceof will not work since the parameter can not be 'dynamic'.

You could use the isInstance method of Class

void foo (Object o)
{
    // Check that o is instanceof a runtime class
    assert getClass().isInstance(o);
}

but this has at least one possible problem:
it will result in an AssertionException if this is an instance of a subclass of A and the object is just a direct instance of A!

edited Jun 18, 2012 at 15:39

answered Jun 18, 2012 at 15:22

user85421

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Comments

Vladislav Bauer · Accepted Answer · 2012-06-18 15:17:52Z

0

In Java, you can check the class of an object using the instanceof operator: object instanceof M

When you have a Class object, it would make sense to write: object.getClass().isAssignableTo(MyClass.class)

http://www.ralfebert.de/blog/java/isassignablefrom/

answered Jun 18, 2012 at 15:17

Vladislav Bauer

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