How wrapper objects work?

Question

Code Snippet

public class WrapperClass {

    public static void main(String[] args) {
        Integer i1 = 400;
        Integer i2 = i1;
        i1++;
        System.out.println(i1 + "  " + i2);
    }
}

Output is 401 400 . I am not sure how wrapper objects work. Arent i1 and i2 pointing to same object ? What happens on java heap when the above code executes ?

[check this question](http://stackoverflow.com/questions/6166348/are-all-primitive-wrapper-classes-immutable-objects) — SomeJavaGuy, Mar 26 '15 at 06:47
What else did you expect? `401 401`? Please read the Java documents... — Afzaal Ahmad Zeeshan, Mar 26 '15 at 06:49

score 3 · Accepted Answer · answered Mar 26 '15 at 06:50

The reason is simple, Wrapper classes are immutable. To explain in detail:-

Integer i1 = 400;
Integer i2 = i1;

Now i1 and i2 point to the same object.

with this i1++ , a new object (with value 401) is created and assigned to i1, while i2 still continues to point to the old object (with value 400).

score 1 · Answer 2 · answered Mar 26 '15 at 06:48

1

The output is correct. Reason behind this is Integer immutability.

answered Mar 26 '15 at 06:48

G. Demecki

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That does not answer most of his question. – Thomas Stets Mar 26 '15 at 06:51
@ThomasStets probably yes, but is more than enough to find out a missing part. – G. Demecki Mar 26 '15 at 06:56

score 0 · Answer 3 · answered Mar 26 '15 at 06:49

0

Yes, of course all wrapper classes are immutable.

answered Mar 26 '15 at 06:49

Viswanath Donthi

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score 0 · Answer 4 · edited Mar 26 '15 at 07:51

0

It is not very hard to understand

i1++;

means

i1 = new Integer(i1.intValue()+1);

edited Mar 26 '15 at 07:51

Mustafa sabir

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answered Mar 26 '15 at 06:49

Mikhail Kasparyan

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What's hard to understand is how you'd write your own wrapper class to do that. It's not like java lets you overload operators. – candied_orange Mar 26 '15 at 06:55

How wrapper objects work?

4 Answers4