overriding

[Shiva Mohan]

jiris mock question

    
public class Chapter8 extends Super {
        
        int x = 20;                             //1
    
        public void test() {
            System.out.println("x = " + x);
        }
    
        public static void main(String[] arg) {
            Super s = new Chapter8();               //2 
            s.test();                           //3
        }
    }
    
    class Super {
        int x = 10;                             //4
    
        public void test() {
            System.out.println("x = " + x);
        }
    } 
    

i was expecting the answer x=10 as member variables are shadowed.so it only takes access of reference variable type.but it printed 20.
the output really hurts my brain.

[Keith Lynn]

The reference to x in Chapter8 is by its simple name, so it refers to the instance variable x in Chapter8 in this case.

[megha joshi]

The reference to x in Chapter8 is by its simple name, so it refers to the instance variable x in Chapter8 in this case.

I agree with that, but what would I write if I needed super's x.

Would it be something like this in Chapter8
int x = super.x

Is there any other way?

[Keith Lynn]

Yes you would need to either use the keyword super or have a getter method in the superclass.

[abi hari]

hi keith,

what does it mean by 'referring to the variable by its simple name'? if this is the case, then in what kind of situation does varible shadowing occur?

Thanks in advance.

[Keith Lynn]

A simple name is a single identifier, like x.

A qualified name is an expression followed by a dot followed by a single identifier, like super.x.

[raj malhotra]

HI
In the above program ans x=20 is correct.

Super s = new Chapter8();               

In the above line as we know that at runtime because of overriding chapter 8 test method will run.As s reference variable contains the reference of Chatpter8 object.so when this program will run x=20 will shadow the x=10 from superclass and we will get x=20 only.

[m ali]

Thanks guys,

   
public class Chapter8 extends Super {
        
        int x = 20;                             //1
 
        public void test() {
            System.out.println("x = " + x);     //how to call Supreme's x? 
        }
    
        public static void main(String[] arg) {
            Super s = new Chapter8();               //2 
            s.test();                           //3
        }
    }
    
    class Super extends Supreme {
        int x = 10;                             //4
    
        public void test() {
            System.out.println("x = " + x);
        }
}
 
 class Supreme {
        int x = 5;                             //5
    
        public void test() {
            System.out.println("x = " + x);
        }
    } 
 

Can anybody explain me the way how to access the 'x' at line5 using Chapter8 object only? I tried it with 'super.super.x' but the compiler slapped me.

Thanks in advance.

[Keith Lynn]

Basically, you can't do it directly. You would have to use a getter method.
[ April 19, 2007: Message edited by: Keith Lynn ]

[Richard Boren]

Originally posted by: Shiva Mohan

public class Chapter8 extends Super {
   int x = 20;                             //1
   public void test() {
      System.out.println("x = " + x);
   } 
   public static void main(String[] arg) {
      Super s = new Chapter8();               //2 
      s.test();                           //3  
   }
}
class Super {      
  int x = 10;                             //4  
  public void test() {
     System.out.println("x = " + x);  
  }
} 

Sorry about reposting of the code, but couldn't think of a better way.

Is x being shadowed? Both the x in Chapter8 and in Super are instance variables so don't they both have the same scope? From K&B 1.5 p.256; "Showding occurs when two variables with different scopes share the same name."

Richard

[Keith Lynn]

That's right.

The declaration of x in the superclass is hidden by the declaration of x in the subclass.