Why is the object not being updated

Question

I have written this code and I wanted to know why the value of the object being referenced is not being changed

All calls in java are call by value. But when the call refers to the same object , why is the object not being updated

class Main {
  public static void main(String[] args) {
    Integer n = 3;
    triple(n);
    System.out.println("Hello world! " + n);
  }
  public static void triple(Integer x)
  {
    x *= 8;
    System.out.println("Hello world! " + x);
  }
}

Actual output

Hello world! 24
Hello world! 3

But I thought that output should be

Hello world! 24
Hello world! 24

Is it so because of the unboxing and autoboxing for the Integer class that a new object is created with the same name as 'x', because Integer is immutable that is locally available and which does not point to the argument/ the object being passed - n.

It outputs `24` because you change the reference locally. The local variable changes, but not the original that was passed in by the `main` method. — markspace, Mar 27 '19 at 04:49
Also, `Integer` is immutable, so of course it's not updated. — markspace, Mar 27 '19 at 04:49

score 2 · Accepted Answer · answered Mar 27 '19 at 05:02

When you pass the Integer n to the triple method you are actually passing the value of the reference of the Integer object n.

So in the triple method another local variable x is pointing to the this reference value when it gets called. Inside the method when it multiplies the value of this Integer object with 8 it will create another new instance of the Integer as Integer is immutable to which the local variable x will point to.

So you won't see this new value in the print statement System.out.println("Hello world! " + n); as n is still pointing to the old instance of the Integer which is still 3.

If you try this with a StringBuilder you would get the result you are expecting:

public static void main(String[] args) {
    StringBuilder n = new StringBuilder("foo");
    triple(n);
    System.out.println("Hello world! " + n);
}

public static void triple(StringBuilder s) {
    s.append("bar");
    System.out.println("Hello world! " + s);
}

Here the output will be:

Hello world! foobar
Hello world! foobar

This is because StringBuilder is mutable, if triple is modifying the data by appending to it, both the original pointer n and the new pointer s will point to the same value of the object reference.

So when I try to modify `x`, a new object is created. Accessible locally. — Nemaly Praveen, Mar 27 '19 at 05:15
@NemalyPraveen yes, as `Integer` is immutable any modification made to an instance will result in a new instance. — Fullstack Guy, Mar 27 '19 at 05:23

Why is the object not being updated

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