Java: Integer equality check

Question

Case 1:

Integer number = 128;
Integer number3 = 128;
System.out.println(number == number3);

Case 2:

Integer number = 127;
Integer number3 = 127;
System.out.println(number == number3);

In this case if the value > 127 it returns false but if the value < 128 then it returns true.

Can somebody explain why case 1 returns false and case 2 returns true?

That's because the values under 127 being cached and using same location of storage for performance reasons in micro level. That's the reason for true in second case. — Suresh Atta, Aug 27 '15 at 10:28
This is my favorite question after "How to compare Strings in Java?" :) — Konstantin Yovkov, Aug 27 '15 at 10:28
@kocko - Don't lie. Your favorite question is *is java pass-by-value or pass-by-reference*? :P — TheLostMind, Aug 27 '15 at 10:32

score 1 · Accepted Answer · answered Aug 27 '15 at 10:27

1

This happens because Integer class keeps a local cache for values between -128 to 127. You could increase this range as well.

answered Aug 27 '15 at 10:27

TheLostMind

it becomes 127 due to 0, zero is taken as a plus value int supports 256. – Kandy Aug 27 '15 at 10:31
@Kandy - Umm.. Nope. Its just a cache – TheLostMind Aug 27 '15 at 10:32
2

It also should be taken into account, that cache works only for wrapping and `valueOf()` declaration. In opposite, `new Integer(5) == new Integer(5)` would return false. – Alex Salauyou Aug 27 '15 at 10:34
1

This is why if you're not sure where exactly these objects come from, use `equals()`, `compareTo()` or explicit cast. – Alex Salauyou Aug 27 '15 at 10:45
@SashaSalauyou - I agree its always good to use `equals()` to compare objects – TheLostMind Aug 27 '15 at 10:47

1 Answers1