2. Given the following code, what will be the output?

2. Given the following code, what will be the output?

Choices:
a. 15 0
    20
b. 15 0
    15
c. 20 0
   20
d. 0 15
   20

Answers:

A is correct.
When we pass references in Java what actually gets passed is the value of that reference (i.e. memory address of the object being referenced and not the actual object referenced by that reference) and it gets passed as value (i.e a copy of the reference is made).
Now when we make changes to the object referenced by that reference it reflects on that object even outside of the method being called but any changes made to the reference itself is not reflected on that reference outside of the method which is called.
 In the example above when the reference v is passed from method first() to second() the value of v is passed.
When we assign the value val to v it is valid only inside the method second() and thus inside the method second() what gets printed is 15 (initial value of i in the object referenced by val), then a blank space and then 0 (value of local variable i).

After this when we return to the method first() v actually refers to the same object to which it was referring before the method second() was called, but one thing should be noted here that the value of i in that object (referred by v inside the method first()) was changed to 20 in the method second() and this change does reflect even outside the method second(), hence 20 gets printed in the method first().
Thus overall output of the code in consideration is 15 0 20