i++ still working for immutable Integer in Java?

Question

I know Integer is immutable in Java. But I tried this:

Integer i = 4;
i++;
System.out.println(i); // output is 5

Why the self-increment is still working? Did Java create a new Integer object?

You've answered the question yourself. – jlordo Jun 03 '13 at 18:46 — jlordo, Jun 03 '13 at 18:46
All are mutable in Java except string. – Rakesh L May 07 '21 at 11:27 — Rakesh L, May 07 '21 at 11:27

score 4 · Accepted Answer · edited Jun 03 '13 at 18:47

4

i++;

is equivalent to:

i = i + 1;

which i now refers to a different Integer object (5).

edited Jun 03 '13 at 18:47

fge

answered Jun 03 '13 at 18:46

Eng.Fouad

`i++` is NOT equivalent to that. `i++` returns the original value, not the new value. – Richard J. Ross III Jun 03 '13 at 18:47
@RichardJ.RossIII can you ealaborate it's quite confusing.... – pinkpanther Jun 03 '13 at 18:49
@RichardJ.RossIII Internally, both compile to `i = Integer.valueOf(i.intValue() + 1)`, so it is actually a different object. Calls to `hashCode()` both before and after the incrementation can verify this as well. – FThompson Jun 03 '13 at 18:50
@pinkpanther the thing about `i++` is that it's a post-increment, meaning that the value changes *after* the statement it's included in executes. – Richard J. Ross III Jun 03 '13 at 18:50
1

@Vulcan It's important to note the side effects of `i++`, however, making this post misleading. – Richard J. Ross III Jun 03 '13 at 18:51
1

@Vulcan - I would think (hope) that hashCode would return a value based on the value of the number, whether immutable or not. – Hot Licks Jun 03 '13 at 18:56

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