[ACCEPTED]-Java FindBugs: Suspicious comparison of Long references-findbugs

Accepted answer

Score: 15

Is this really an issue and if so, how do 7 I fix it?

Yes, you're comparing references not values, so 6 two different Long instances with identical 5 values will result in that comparison being 4 false. You should use Long#equals() instead:

final Long id1 = materialDefinition.getId();
final Long id2 = definitionToRemoveFromClass.getId();

if (id1.equals(id2)) {
    //...
}

If getId() can return 3 null, then also be sure to include the appropriate 2 null check (although getId() doesn't sound like a method 1 that should be returning null).

Score: 1

You may want to do something like this.

public static boolean equals(Long a, Long b){
    if (a==null && b==null) return true;
    else if ((a==null) != (b==null)) return false;
    else return a.equals(b);
}

Score: 0

You'd better use if (materialDefinition.getId().equals(definitionToRemoveFromClass.getId()))

Be sure that materialDefinition.getId() is not null.

To 2 compare these values (< >) you can use 1 .compareTo() method :

A == B: A.compareTo(B) == 0

A < B: A.compareTo(B) < 0

A > B: A.compareTo(B) > 0

Score: 0

Yes it's a bad smell, and you should use 11 the equals() method instead, as explained @arshajii 10 above.

What I've found most intriguing is 9 the fact that such comparison might even 8 work sometimes, like starting with small 7 values (-128 to 127), and will break for 6 larger numbers. For example:

@Test
 public void compareSmallIds() {
     Long id1 = 127L;
     Long id2 = 127L;

     Assert.assertTrue(id1 == id2);
     Assert.assertTrue(id1.equals(id2));
     Assert.assertTrue((long) id1 == (long) id2);
     Assert.assertTrue(id1.longValue() == id2.longValue());
 }

@Test
 public void compareIds() {
     Long id1 = 500L;
     Long id2 = 500L;

     Assert.assertFalse(id1 == id2);
     Assert.assertTrue(id1.equals(id2));
     Assert.assertTrue((long) id1 == (long) id2);
     Assert.assertTrue(id1.longValue() == id2.longValue());
}

It seems this 5 behavior is based on number cache policy 4 definition, with byte values range as default. But 3 definitely, code with non-primitive values 2 should not rely on the == operator for value 1 comparison!

Source: stackoverflow.com

[ACCEPTED]-Java FindBugs: Suspicious comparison of Long references-findbugs

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