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What will be the output of the program? class Two { byte x; } class PassO { public static void main(String args) { PassO p = new PassO(); p.start(); } void start() { Two t = new Two(); System.out.print(t.x + " "); Two t2 = fix

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1. 

What will be the output of the program?

class Two
{
byte x;
}

class PassO
{
public static void main(String [] args)
{
PassO p = new PassO();
p.start();
}

void start()
{
Two t = new Two();
System.out.print(t.x + " ");
Two t2 = fix(t);
System.out.println(t.x + " " + t2.x);
}

Two fix(Two tt)
{
tt.x = 42;
return tt;
}
}

[A]. null null 42
[B]. 0 0 42
[C]. 0 42 42
[D]. 0 0 0

Answer: Option C

Explanation:

In the fix() method, the reference variable tt refers to the same object (class Two) as the t reference variable. Updating tt.x in the fix() method updates t.x (they are one in the same object). Remember also that the instance variable x in the Two class is initialized to 0.

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