Problem with output( Casting)

[Rohan Deshmkh]

This is a question from enthuware's.

interface I{
}
class A implements I{
}

class B extends A {
}

class C extends B{
}

And the following declarations:
A a = new A();
B b = new B();
Identify options that will compile and run without error.


1. a = (B)(I)b;
2. b = (B)(I) a;
3. a = (I) b;
4. I i = (C) a;

I am not able to understand it's output.The correct answer is option 1.How? I assumed it is correct because b is an instance of B.No matter what's on the left side of = and not taking into consideration about(I) in the expression (B)(I)b

Option 2 is not correct beacuse a is not an instance of B.

Why is option 3 not correct.b is an instance of I(Indirectly).Look's like now my assumption of not looking on the left side of = is false and now it does matter to whom you are assigning the variable.
Option 4 is wrong because a is a A but A is not C.So we cannot cast this way.


Am i correct?

[Jesper de Jong]

Option 3 is wrong because you cannot assign any arbitrary I to a variable of type A.

Class A implements interface I, but there could be other classes, unrelated to A, that also implement I. If option 3 were possible, you could do:

class D implements I { ... }

D d = new D();

a = (I)d; // This should not work!

[harshvardhan ojha]

Hi Rohan, why you are assuming that left side of = doesn't matter? Although your other assumptions seems to be correct only look at option C, you need to make your understanding about how a subclass casted to interface cant be assigned to superclass. Again superclass is not subclass.

[Tony Docherty]

1. a = (B)(I)b;

Type B is upcast to I - Legal as B extends A which implements I
Type I is downcast to B - Legal as the object in this case is of type B
Type B is assigned to variable of type A - Legal, you can always assigned an object to a variable which is a super type. ie it is always legal to upcast.

2. b = (B)(I) a;

Type A is upcast to I - Legal as A implements I
Type I is downcast to B - Illegal as the object in this case is of type A and can't be downcast to type B

3. a = (I) b;

Type B is upcast to I - Legal as B extends A which implements I
Type I is assigned to variable of type A - Illegal, you can't downcast unless you specify the type you are downcasting to.

4. I i = (C) a;

Type A is downcast to C - Illegal as the object in this case is of type a and can not be downcast to type C.

.Look's like now my assumption of not looking on the left side of = is false

Yes you are right in realising your assumption is wrong. See the answer in your other thread.

[Rohan Deshmkh]

Ok , i finally got it.
Thanks to everyone , especially Tony Docherty for nice and step by step explanation.