reference casting

[Brian Deer]

Any help with reference variable casting is much appreciated.

Here is the question from K&B that threw me off:

Given:
class Putter{
public static void main(String[]args){
Bango b1=new Bango();
Bango b2=new Bongo();
Bango b3=new Bingo();
//insert code
}
}
class Bango{}
class Bongo extends Bango{}
class Bingo extends Bongo{}
Which, inserted at "//insert code", will NOT compile?(Choose all that apply.)

A.Bongo b4=b2
B.Bongo b5=(Bongo)b2
C.Bango b6=b3
D.Bango b7=(Bango)b3
E.Bingo b8=b3
F.Bingo b9=(Bingo)b3
G.Bango b10=(Bongo)b3

Answer: A and E will not compile.

For one, I guess I don't understand how G can work.

Secondly, Is there a shortcut or some other way to think of it that may help me wrap my mind around the concept?

Help.
thanks,
brian

[Nik Arora]

Hi Brian,

Any help with reference variable casting is much appreciated.

Here is the question from K&B that threw me off:

Given:
class Putter{
public static void main(String[]args){
Bango b1=new Bango();
Bango b2=new Bongo();
Bango b3=new Bingo();
//insert code
}
}
class Bango{}
class Bongo extends Bango{}
class Bingo extends Bongo{}
Which, inserted at "//insert code", will NOT compile?(Choose all that apply.)

A.Bongo b4=b2
B.Bongo b5=(Bongo)b2
C.Bango b6=b3
D.Bango b7=(Bango)b3
E.Bingo b8=b3
F.Bingo b9=(Bingo)b3
G.Bango b10=(Bongo)b3

Answer: A and E will not compile.

For one, I guess I don't understand how G can work.

Secondly, Is there a shortcut or some other way to think of it that may help me wrap my mind around the concept?

Remember this : During compile time compiler checks on reference variable type and during run-time its based on object type.

Bango b3=new Bingo();

G.Bango b10=(Bongo)b3

b3 is a reference type of Bango and the object is of Bingo.In the code Bongo class extends Bango. So during compile time the cast passes the IS-A test. So the cast will compile.


Regards
NIK
SCJP 1.5

[Anthony O'Dea]

Brian

class Putter{
public static void main(String[]args){
Bango b1=new Bango();
Bango b2=new Bongo();
Bango b3=new Bingo();
//insert code
}
}
class Bango{}
class Bongo extends Bango{}
class Bingo extends Bongo{}

Which, inserted at "//insert code", will NOT compile?(Choose all that apply.)

A.Bongo b4=b2
B.Bongo b5=(Bongo)b2
C.Bango b6=b3
D.Bango b7=(Bango)b3
E.Bingo b8=b3
F.Bingo b9=(Bingo)b3
G.Bango b10=(Bongo)b3

Answer: A and E will not compile.

For one, I guess I don't understand how G can work.

Secondly, Is there a shortcut or some other way to think of it that may help me wrap my mind around the concept?

Technically, all of the answers are correct since the question asks which WILL NOT COMPILE, and the semi-colon is missing for all of them

G doesn't work, but it does COMPILE (as Nik has said).

Simply put, anything with an explicit cast will COMPILE (but may throw an exception at runtime) - provided the cast type and the reference type are in the same hierarchy. So B, D, F and G all compile.

If the reference type of the object being assigned does not match or is not a subclass of the reference type being assigned to, it will not compile without a cast. So A (assigning a Bango to a Bongo) and E (assigning a Bango to a Bingo) do not compile, as they are assigning a superclass to a subclass.

[Amol Nayak]

@ Anthony

Why wont G work?

its perfectly legal at runtime as well.

The object in b3 is of Bingo and can be cast to Bongo as Bongo is Superclass of Bingo...

[anil kumar]

Hi

--------------------------------------------------------------------------
Bango b10=(Bongo)b3
--------------------------------------------------------------------------

The meaning of the above statement is this.
Bongo b=(Bongo)b3;

Bango b1=(Bango)b;

Think about this ,if you think the above statement will work then the below statement will also work.
Bango b10=(Bongo)b3.



Thanks

Anil Kumar

[Brian Deer]

Thank you all for your responses. Sorry I left the semi-colons out of the second half of the example.

brian