Question

1. Given the following code, how many references exist to the Time object at the end of the main method and what are their names? public class Foo { public static void main(String args[]) { Time t = new Time(); Time t1 = t; Time t2 = t; Clock clock = new Clock(t); // AT THIS POINT, HOW MANY REFERENCES TO THE TIME OBJECT ARE THERE } } public class Time { } public class Clock { private Time time; public Clock(Time t) { time = t; } }

2. What does the following code display to the screen? public void run() { int num1 = 20; int numArray[] = {30,40,50,60,70}; MyMethod(num1); System.out.println(num1); MyMethod(numArray); System.out.println(numArray[2]); } public void MyMethod(int i) { i += 5; } public void MyMethod(int i[]) { for (int j = 0; j < i.length; j++) { i[j] += 5; } }

Answer 1

I have answered both the questions then please leave a comment I will get back to you.

1. In this program there is Only One object i.e. t and three references that is:- t, t1 and t2
       where

• t points to new Time object in memory (line Time t = new Time();)
• t1 will refer to time object created by t shown by line : Time t1 = t;
• t2 will refer to time object created by t shown by line :   Time t2 = t;

2. Nothing will be displayed if the code is run as it is. As for java program the starting point is main()

Is you update the code and call make run from main() then the output would be:

20
55

and the logic behind this that first print statement is just printing the num1 which is not updated as in java arguments are always passed by value so num1 will not be updated after the call MyMethod(num1); and System.out.println(num1); will print 20

And in case of array this reference will be passed by value, which is copied. It will still point at the original array. So, References to objects are passed by value and hence the value updated in method MyMethod(int i[]) will be depicted in the print statement 55