Question

What is printed by running the following code?

public static void main(String[] args) {

int[] nums = {2, 3, 4};

int n = 5;

changeMe1(n, nums);

System.out.print( n );

System.out.print(nums[0]);

changeMe2(n, nums);

System.out.print( n );

System.out.print(nums[0]);

}

public static void changeMe1(int number, int[] list) {

number++;

list[0]++;

}

public static void changeMe2(int number, int[] list) {

number = 9;

list = new int[1];

list[0] = 99;

}

Answer 1

class Outpt
{

public static void main(String[] args) {

int[] nums = {2, 3, 4};

int n = 5;

changeMe1(n, nums);

System.out.print( n );   // n still remains same after function call

System.out.print(nums[0]);

changeMe2(n, nums);

System.out.print( n ); // n still remains same after function call

System.out.print(nums[0]);

}

public static void changeMe1(int number, int[] list) {

number++;   //value of local variable number incremented but not reflected in variable n in main since it is a pass by value call for changeMe1

list[0]++; ////array is passed by reference and first value incremented and is reflected in original array nums in main

}

public static void changeMe2(int number, int[] list) {

number = 9; //value of local variable number changed to 9 but not reflected in variable n in main since it is a pass by value call for changeMe1

list = new int[1]; //array is passed by reference but here it is locally defined as a single variable array and given value 99 which is not reflected in original array nums in main

list[0] = 99;

}
}

// Thus the output of program is : 5353