Java passByValue or Reference confusion

Question

I was trying to play with Java references and I got an interesting situation. This following piece of code gives unpredictable output to me. I am trying to modify an array, a string and an integer in a function.

 public static void main(String[] args){
     int[] arr = {1,2,3,4,5}; 
     Integer b = 6;      
     String s = "ABC";

     fun(arr, b,s);
     for(int i : arr)
         System.out.print(i + " ");
     System.out.println();
     System.out.println("b="+b);
     System.out.println("s="+s);
 }   

 public static void fun(int[] a, Integer b, String s){
     b = b*10;       
     for(int i =0; i<a.length; i++)
     {
         a[i] = a[i]+10;
     }       
     s=s+"PIY";
 }

Now this gives following output :

11 12 13 14 15 
b=6
s=ABC

I don't understand why array is getting changed but string and integers are not getting changed inside the function.

Notice there's a `b=` and `s=` but no `a=`? That's because you're modifying `a`, not reassigning it. — shmosel, Aug 24 '17 at 05:22

score 3 · Answer 1 · answered Aug 24 '17 at 05:11

3

Array is an Object and Integer and String are immutable in Java. You cannot change immutable object by reference. You have to reinsert/reassign to see the changes. Hence the difference.

Your logic applies and correct in case of general objects which are not immutable

answered Aug 24 '17 at 05:11

Suresh Atta

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Thanks But what do you mean by reassign. For me, I already reassign inside the function . – thatman Aug 24 '17 at 05:18
@Piyush_Chhabra Reassign to the array position. Not to object. – Suresh Atta Aug 24 '17 at 06:17

Java passByValue or Reference confusion

1 Answers1