Java OCA OCP Practice Question 683

Question

What will the following program print when run? 

public class Main  { 

    private int myValue = 0; 
     //  w w  w. j  a v a2s  .  c o  m
    public void showOne (int myValue){ 
        myValue = myValue; 
     } 
     
    public void showTwo (int myValue){ 
        this.myValue = myValue; 
     }     
    public static void main (String [] args)  { 
        Main ct = new Main (); 
        ct.showTwo (200); 
        System.out.println (ct.myValue); 
        ct.showOne (100); 
        System.out.println (ct.myValue); 
     } 
}

Select 1 option

  • A. 0 followed by 100.
  • B. 100 followed by 100.
  • C. 0 followed by 200.
  • D. 100 followed by 200.
  • E. 200 followed by 200.
  • F. 200 followed by 100 


Correct Option is  : E

Note

Within an instance method, you can access the current object of the same class using 'this'.

When you access this.myValue, you are accessing the instance member myValue of the Main instance.

If you declare a local variable (or a method parameter) with the same name as the instance field name, the local variable "shadows" the member field.

You should be able to access the member field in the method directly by using the name of the member (in this example, myValue).

Because of shadowing, when you use myValue, it refers to the local variable instead of the instance field.

In showTwo() method, there are two variables accessible with the same name myValue.

One is the method parameter and another is the member field of Main object.

You should be able to access the member field in the method directly by using the name myValue but in this case, the method parameter shadows the member field because it has the same name.

So by doing this.myValue, you are changing the instance variable myValue by assigning it the value contained in local variable myValue, which is 200. So in the next line when you print ct.myValue, it prints 200.

In the showOne() method also, there are two variables accessible with the same name myValue. One is the method parameter and another is the member field of Main object. So when you use myValue, you are actually using the method parameter instead of the member field.

When you do : myValue = myValue; you are actually assigning the value contained in method parameter myValue to itself. You are not changing the member field myValue. Hence, when you do System.out.println(ct.myValue); in the next line, it still prints 200.




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