polymorphism Problem (tricky)

[Mohit G Gupta]

TestCafe by sahir shah

What will it print now?

 
  
   class Base{
      static  int value = 0;
        Base(){
        addValue();
        }
    static void addValue(){
        value += 10;
        }
       int getValue(){
         return value;
       }
     }
    class Derived extends Base{
      Derived(){
       addValue();
      }
    static void addValue(){
     value +=  20;
      }
    }
    public class Test {
      public static void main(String[] args){
          Base b = new Derived();
          System.out.println(b.getValue());
      }
    }
   

output:30

i think output should be 20

 Base b = new Derived();   

will call the Derived class constructor which will in turn call Base class constructor.
now addValue method call as in BAse class Constructor would be called
then,addValue method of Base class would be called as static methods are not overriden and in case of static methods reference type determines which method to call value=10
then addValue method call in Derived class constructor would be called therefore value=10+10=20

[Ram Narayan.M]

Derived object is created, then, Derived() is called and then base class constructor is implicitly called since super() is the first statement of each constructor....

In Base(), addValue() of Base class is called, value is incremented to 10...

After Base() constructor initialization, then Derived(), addValue() of Derived class is called, value is incremented to 30...

Note : static value is shared by Derived class from Base class due to inheritance...

[Abimaran Kugathasan]

mohitkumar gupta wrote:
then,addValue method of Base class would be called as static methods are not overriden and in case of static methods reference type determines which method to call value=10
then addValue method call in Derived class constructor would be called therefore value=10+10=20

Those are different stuffs. When you are creating the Derived class object, the compiler doesn't care about, to what reference type variable you are going to assign this object. Assigning to a super type, doesn't make any impact on the creation of the object. Only after creating the object, you are going to assign it.

And, make the method as instance methods, and predict the output. It should be 40.

[Mohit G Gupta]

Then,what about this question

   
   class Base{  
        int value = 0;  
        Base(){  
        addValue(); 	System.out.println("const of base"); 
        }  
     void addValue(){ 
     	System.out.println("addValue base");  
        value += 10; 
         
        }  
       int getValue(){  
         return value;  
       }  
     }  
    class Derived extends Base{  
      Derived(){  
       addValue();  
       	
        	System.out.println("const of Derived");
      }  
     void addValue(){  
     	
     	System.out.println("addValue Derived");
     value +=  20;  
      }  
    }  
    public class Testp {  
      public static void main(String[] args){  
          Base b = new Derived();  
          System.out.println(b.getValue());  
      }  
    }   

output:40


here Derived class addVaue method is called twice,Base class addValue is not callled even once

why ???

[Abimaran Kugathasan]

mohitkumar gupta wrote:
here Derived class addVaue method is called twice,Base class addValue is not callled even once

why ???

That's because of VMI and the actual object is Derived But, a kind advice : Don't invoke overridden methods in Constructors.

[Shanky Sohar]

In case of overridden method object type determines which method will call.

and object is of derived not a base so addValue of derived class is called

[Shanky Sohar]

And you have already put the sys.out so i think you understand how the flow of the code is going on.