Why comparing two longs with == returns false and when I cast to int or String it returns true?

Question

I'm doing the following comparison:

atividade.getEscala().getId() == escala.getId()

and it returns false but if I type atividade.getEscala().getId().intValue() == escala.getId().intValue() then it returns true. It also returns true when I write atividade.getEscala().getId().toString().equals(escala.getId().toString())

I know by debugging that the content of both variables is the same ( (java.lang.Long) 2 in display view ), then why does it returns false when I compare the longs with just ==?

Integers have an internal pool for the lowest values, I'm not sure about longs. Your string question doesn't make too much sense — Jeroen Vannevel, Jul 25 '14 at 18:54
I edited because I had written the equals comparison wrong. it also return true when I use equals. — StudioWorks, Jul 25 '14 at 18:56
`Long` (capital) is a wrapper class, so it's an object and you're comparing references — Ben, Jul 25 '14 at 18:56
http://stackoverflow.com/questions/20541636/compare-non-primitive-data-type-long-values-in-java — Andreas, Jul 25 '14 at 18:57

Peter Lawrey · Accepted Answer · 2014-07-25T19:36:58.493

When you use == with two references, you are comparing whether the references are the same, not whether the contents of the objects referenced are the same.

This is made more complicated with auto-boxing caches. e.g. Longs between -128 and 127 are cached so you will get the same object every time.

e.g.

Long x = 100, y = 100;
System.out.println(x == y); // true, references are the same.

Long a = 200, b = 200; // values not cached.
System.out.println(a == b); // false, references are not the same.
System.out.println(a.equals(b)); // true, the objects referenced are equal

Comparing the intValue() is dangerous due to the risk of overflow.

Long c = -1, d = Long.MAX_VALUE;
System.out.println(c.equals(d)); // false
System.out.println(c.longValue() == d.longValue()); // false
System.out.println(c.intValue() == d.intValue()); // true, lower 32-bits are the same.

I see, my id is a `Long` not a `long`. Shoul I just use `.intValue()` then? — StudioWorks, Jul 25 '14 at 18:56
@StudioWorks best to use `.equals()` or if you have to `.longValue()` as two longs can be different but have the same `.intValue()` — Peter Lawrey, Jul 25 '14 at 19:37

score 3 · Answer 2 · answered Jul 25 '14 at 18:56

The == operator compares identity, not equality. For equality, use equals().

identity: Two distinct variables share a common object. In other words, they point to the same underlying object, and thus reference the same memory address.
equality: Two distinct objects represent the same thing. In other words, they are different objects, but have the same contents.

score 1 · Answer 3 · answered Jul 25 '14 at 18:56

1

When you use == operator on objects you compare memory reference values, not values. For primitives you compare values.

answered Jul 25 '14 at 18:56

alobodzk

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score 1 · Answer 4 · answered Jul 25 '14 at 18:56

1

If both values are Long, you need to compare them with equals(), as you do for all reference types. By calling intValue(), you're comparing int primatives, for which == works as you'd expect.

answered Jul 25 '14 at 18:56

yshavit

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Why comparing two longs with == returns false and when I cast to int or String it returns true?

4 Answers4