Why (Integer) 222 != (Integer) 222 in Java?

Question

It holds true for (Integer) 1 == (Integer) 1, which seems legitimate.

So why it's having excursion for (Integer) 222's equality?

Just don't use == "because it works for this case", you are comparing objects, so use equals. — Viktor Mellgren, Feb 15 '16 at 12:37

Davide Lorenzo MARINO · Accepted Answer · 2016-02-15T12:41:08.750

9

Integer is a class. So to compare objects you need to use equals instead of ==

What actually happens with shorter Integer is that if you get an Integer using the method valueOf you get always the same cached instance for values between -128 and 127. So in this case == works.

It doesn't work if you instead of using valueOf create a new instance explicitly with the operator new.

For To be more clear I write the current implementation of valueOf

public static Integer valueOf(int i) {
    final int offset = 128;
    if (i >= -128 && i <= 127) { // must cache
        return IntegerCache.cache[i + offset];
    }
    return new Integer(i);
}

edited Feb 15 '16 at 12:41

answered Feb 15 '16 at 12:32

Davide Lorenzo MARINO

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but it holds true for (Integer) 1 == (Integer) 1 ? – Snehal Masne Feb 15 '16 at 12:34
Check for the updated answer – Davide Lorenzo MARINO Feb 15 '16 at 12:36
Interesting -1 for a correct answer – Davide Lorenzo MARINO Feb 15 '16 at 12:37
I didn't downvote, but people also downvote for people answering questions that should not be answered - in this case because it is an obvious duplicate. – Gimby Feb 15 '16 at 12:45
5

Yes but it is always a bad behaviour to downvote a correct answer. Better is to vote to close the question, not to downvote a correct answer. – Davide Lorenzo MARINO Feb 15 '16 at 12:46

Why (Integer) 222 != (Integer) 222 in Java?

1 Answers1