OverLoading && Overrriding confused with output

[Ganesh Gowtham]

Hi All,

 package scjpsamples;
 
class Az
{
void m1(Az a)
{
System.out.println("A");
}
void dep()
{
System.out.println("dep in A");
}
}
class Bz extends Az
{
void m1(Bz b)
{
System.out.println("B");
}
void dep()
{
System.out.println("dep in B");
}
}
class Cz extends Bz
{
void m1(Cz c)
{
System.out.println("C");
}
void dep()
{
System.out.println("dep in C");
}
}
class InvokeEx
{
public static void main(String[] args)
{
Az c1 = new Cz();
Cz c2 = new Cz();
[B]c1.m1(c2);// prints A[/B] //line 1
[B]c2.m1(c2);// prints C[/B]  //line 2
c1.dep();// ok for me 
}
}   

i am Ok with output displayed for c1.dep() (overridden) ,
Could any one please explain me how exactly we get output in as 'A' and 'C' in line 1 and line 2 .

[dhwani mathur]

Hi Gowtham!!

  class InvokeEx
{
public static void main(String[] args)
{
Az c1 = new Cz();
        Cz c2 = new Cz();
        c1.m1(c2);// prints A //line 1
                c2.m1(c2);// prints C  //line 2
                c1.dep();// ok for me 
}
}   
 
this is code shown by you,one thing i wil like to say you is
when we pass object as reference to method then the method 
of the class whoes reference it holds will be called,as shown
Az c1=new Cz();
c1.m1(c2)
 
Here c1 is reference variable of class Az ,so method from class Az will be called
so now when we //c1.m1(c2)
call method m1( with c2 as argument) ,although c2 is object of class Cz but c2 becomes 
a subtype of Az [B]bcoz Cz is a subclass of Az[/B],so it is acceptable
to pass c2 as argument......Hence it prints A
 
Now comes  next case 
c2.m1(c2)
 
here you can see that both reference variable as well as
argument is c2 which is of type Cz so it prints C
  

I hope this helps you to atleast some extent..............
[ August 20, 2007: Message edited by: dhwani mathur ]

[Ganesh Gowtham]

Hi mathur
Thanks for your reply .
see the below code .

  package scjpsamples;
 
class Az
{
void m1(Az a)
{
System.out.println("A");
}
void m1(Cz aa)
{
System.out.println("A XXX");
}
void dep()
{
System.out.println("dep in A");
}
}
class Bz extends Az
{
void m1(Bz b)
{
System.out.println("B");
}
void dep()
{
System.out.println("dep in B");
}
}
class Cz extends Bz
{
void m1(Cz c)
{
System.out.println("C");
}
void dep()
{
System.out.println("dep in C");
}
}
class InvokeEx
{
public static void main(String[] args)
{
Az c1 = new Cz();
Cz c2 = new Cz();
c1.m1(c2);// prints A
c2.m1(c2);// prints C
c1.dep();// ok for me 
}
}  

i was bit even more confused .... but upto some extend i will agree with you .
see the output for the this code .i added one more method in class Az.

[Peter Chase]

The revised version doesn't do what you say it does: -

C:\temp>%CV_JDK_1_6_0%\bin\java InvokeEx
%CV_JDK_1_6_0%\bin\java InvokeEx
C
C
dep in C

The original version does do what you say: -

C:\temp>%CV_JDK_1_6_0%\bin\java InvokeEx
%CV_JDK_1_6_0%\bin\java InvokeEx
A
C
dep in C

In the original, there is no overriding of any of the m1(something) methods. The three classes provide three different overloads of it.

The variable c1 is declared to contain an Az object. When choosing between overloads, it is the declared type that determines what happens. There is only one method in Az that matches c1.m1(c2); it's the one that prints "A".

In the modified version, the class Az has a method m1(Cz) that is overridden in class Cz. When choosing between overrides, it is the run-time type that matters. So, because c1's run-time type is Cz, the method in Cz is called, printing "C".
[ August 20, 2007: Message edited by: Peter Chase ]

[Ganesh Gowtham]

Hi Peter ,
Thanks for replying .
Here c1 has object which refer;s to "Cz" class.
what you had mentioned in repy "The variable c1 is declared to contain an Az object" which comfuses me .

Please let me know if am wrong in understanding the concept.?