Question

I do not understand why the swap methods are not doing anything for the variable x and y in the main method. When are variables affected by the other methods and when they do not get modified by the other methods? (The answer is A).

Thank you so much in advance,

24. What does the following code print? public class Swapper [ public static void main (String[] args) I int [ ] x = int [ ] y = {4,5,6,7,8}; swap (x,y); System.out.println (x. 1length+""+ y. length); swap (x [0], y[0]); System.out.println (x[0]""+ y [0]) public static void swap (int[] x, int[] y) f int [ ] temp = x; x = y; = X; public static void swap (int x, int y) [ int temp = x; x=y; (A) 3 5

Answer 1

As we know, in Java whenever we pass parameter in functions it is always call-by-value. Java doesn't support call-by-reference mechanism. So doesn't matter whether we pass  any array swap(x,y) in function ( although array name points to it's base address and in C/C++ if we pass any any array in function then changes will reflect in origial array in main function but it is not true in JAVA ) or any values swap(x[0],y[0]) in function, everything will call-by-value mechanism and it won't affect values of X,Y in main function because copy of original values is being passed in functions not actual variables.