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?
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FThompson
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lakeskysea
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1You've answered the question yourself. – jlordo Jun 03 '13 at 18:46
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All are mutable in Java except string. – Rakesh L May 07 '21 at 11:27
1 Answers
4
i++;
is equivalent to:
i = i + 1;
which i now refers to a different Integer object (5).
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`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
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@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
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@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
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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
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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