constructor

[swapnil paranjape]

please explain the output

i was expecting the output as
a=10
b=16

but output is
b = 0
b = 16

actualy subclass constructor should call super (base class constructor) but thats is not happening here..please explain

  
class bc
{
int a = 10;
 
bc ( )
{
call ( ) ;
}
void call ( )
{
System.out.print( "a = " + a + " ") ;
}
}
 
public class abc extends bc
{
int b = 16 ;
abc() 
{
call ( ) ;
}
 
void call ( ) 
{
System.out.println( " b = " + b + " " ) ;
}
public static void main ( String args [ ] )
{
new abc( ) ;
}
}

[Deepak Bala]

I think this is what you were trying to do...

  
class bc
{
int a = 10;
 
bc ( )
{
call_one ( ) ;
}
void call_one ( )
{
System.out.print( "a = " + a + " ") ;
}
}
 
public class abc extends bc
{
int b = 16 ;
abc() 
{
call ( ) ;
}
 
void call ( ) 
{
System.out.println( " b = " + b + " " ) ;
}
public static void main ( String args [ ] )
{
new abc( ) ;
}
}
  

The function call is over ridden, so the super constructor calls the derived class call() function. At this point of time b is not initialized with 16 so it prints 0. I changed the function name from call to call_one in the base class.

[Flom Xanther]

just for your info,
if you would make call() in bc private, then the result is what you expect..

Because private methods/attributes are not inherited, so no way for polymorphism..

regards,
Flom

[swapnil paranjape]

when the base class contructor calls the method call() its actually this.call() (am i right?) so it should invoke the base class call method. why the call method from subclass is called?

please explain.