Please Explain this Java Array Reference Parameter Passing Behavior
arrays, java, parameter-passing, pass-by-value
Solution
In Java, it's a misnomer to say that objects are passed by reference. It's more accurate to say that the reference to the object is passed by value.
You pass the array reference to `reverseArray` by value. The local parameter is a copy of the reference to the array. Later when you say
ar = temp;
You have only pointed the local `ar` to `temp`, not the original array reference `ar` from `main`.
On the other hand, in the `shiftRight` method, you have directly accessed the array through the copied reference, so the original array's contents change and the method works as expected.
Problem
``` public class TestArray { public static void main(String[] args) { int[] ar = {1,2,3,4,5,6,7,8,9}; shiftRight(ar); for (int i = 0; i < ar.length; i++) { System.out.print(ar[i]); } // prints: 912345678 -- good System.out.println(); reverseArray(ar); for (int i = 0; i < ar.length; i++) { System.out.println(ar[i]); } // prints: 91234567 -- I don't understand System.out.println(); } public static void shiftRight(int[] ar) { int temp = ar[ar.length - 1]; for (int i = ar.length - 1; i > 0; i--) { ar[i] = ar[i - 1]; } ar[0] = temp; } public static void reverseArray(int[] ar) { int[] temp = new int[ar.length]; for (int i = 0, j = temp.length - 1; i < ar.length; i++, j--) { temp[i] = ar[j]; } ar = temp; for (int i = 0; i < ar.length; i++) { System.out.print(ar[i]); } // prints: 876543219 System.out.println(); } } ``` Passing an array to a parameter results in passing the reference to the array to the parameter; if an array parameter is changed within the method, that change will be visible outside of the method. The first method, `shiftRight`, does what I expect it to: it changes the array outside of the method. The second method, however, does not change the array outside of the method. But running the for loop inside of the method prints the correct values. Why isn't the reference of `ar` pointed to `temp`? Is it because the variable `temp` is destroyed when the method stops--does that kill the reference as well? Even if this is the case, why does Java take `ar`, which was pointed to the reference of `temp` and then reapply it the the original reference of `ar`? Thank you.