About Question enthuware.ocajp.i.v7.2.1015 :

[Vermeulen]

Why would this question be marked "Very easy"?

I found it a bit hard because I didn't know that the compiler would stop looking for i when it found the private int B.i. Instead I thought the compiler might skip it because it's private, and use A.i instead.

Of course once you know the compiler does this, the question is not hard to answer.

In several years of programming I've never seen this situation. The reason is that it's not a good (therefore not common) practice to use public variables or (arguably) even protected ones, therefore I have never encountered this situation!

[admin]

Toughness is fairly subjective but I agree that it isn't too easy.
-Paul.

[wuqing1450]

Why would this question be marked "Very easy"?

I found it a bit hard because I didn't know that the compiler would stop looking for i when it found the private int B.i. Instead I thought the compiler might skip it because it's private, and use A.i instead.

Of course once you know the compiler does this, the question is not hard to answer.

In several years of programming I've never seen this situation. The reason is that it's not a good (therefore not common) practice to use public variables or (arguably) even protected ones, therefore I have never encountered this situation!

If the modifiers of A.i and B.i are changed to private and public respectively, then B.i can be used by the compiler and printed 30

Code: Select all

class A {    
       private int i = 10;
       private int j = 20;  
}  
class B extends A {    
       public int i = 30; //1    
       public int k = 40;  
}  
class C extends B  { }  
public class TestClass {    
       public static void main(String args[]) {       
              C c = new C();
              System.out.println(c.i); //2 will print 30
              System.out.println(c.j); //3 compile error
              System.out.println(c.k);    // will print 40
       } 
}  

[shambhavi]

CODE 1 :

Code: Select all

interface A
{
  int i = 10;  
    int j = 20;
}
interface B extends A
{
   int i = 30;    
    public int k = 40;  
}
interface C extends B{ }
      
 class scrap implements B,A { // LINE 1
  public static void main(String[] args) { 
      scrap c = new scrap();
      System.out.println(c.i); // compiler error -  reference to ‘ i ’ is ambiguous 

}  }

CODE 2 :

Code: Select all

interface A
{
  int i = 10;  
    int j = 20;
}
interface B extends A
{
   int i = 30;   
    public int k = 40;  
}
interface C extends B{ }
      
 class scrap implements B{ // LINE 1
  public static void main(String[] args) { 
      scrap c = new scrap();
      System.out.println(c.i); // COMPILES FINE 

}  }

LINE 1 in both the codes are producing different behaviours . in code 1, scrap implements B,A actually means that it implements only B right, since A is redundant , and implementing B will automatically mean implementing A too. but yet , it shows an ambiguity error. why doesn't B's ' i ' hide A here ?

isn't code 2 and and code 1 same ? I mean isn't the LINE 1 in code 1 and 2 actually same i.e. in code 1 saying it implements A is redundant ?

[admin]

That is the just a rule of resolving ambiguity. In the second code, Since scrap implements B explicitly, the compiler is able to make the decision in favor of B's i by giving more weightage to the explicitly implemented interface.
Try the specification to see the specific rule.

[shambhavi]

thanks

[crazymind]

Just need to double confirm that: variable hiding(btw Parent class and Child class) and variable shadowing happens (btw instance variable and local variable) only if two variable name are same (nothing to do with identifier or type?)

variable hiding:

Parent class:
final double a = 1;

Child class:
int a = 3;

Will this a hide a from parents class?

[admin]

Yes, it will. Please try it out.

[crazymind]

Hi, forgive my poorly English reading ability, are "commented" and "commented out" same thing here?

[admin]

Yes, same thing in this case.

[crazymind]

Thanks