Comparing two Integers 'Unnecessary unboxing' warning

Question

I try to compare Integers the next way (for my case, it's good):

public void compareMayNull(Integer a, Integer b) {
    if ((a == null ? -1 : a.intValue()) == b.intValue())
        System.out.println("true");
}

IntelliJ IDEA gives me warning 'Unnecessary unboxing' on a.intValue() and b.intValue() and recommend me to simplify code to this:

public void compareMayNull(Integer a, Integer b) {
    if ((a == null ? -1 : a) == b)
        System.out.println("true");
}

But I'm little confused, because references will be compared if a != null that not best practice as I know. What code should I use?

http://stackoverflow.com/questions/1514910/how-to-properly-compare-two-integers-in-java — Neo, Nov 08 '16 at 08:14
Since you assume that `b` cannot be `null`, how about `if(b.equals(a == null ? -1 : a))` ? — Thilo, Nov 08 '16 at 08:14
If `b` cannot be `null`, you could also make this explicit by changing the signature to `compareMayNull(Integer a, int b)` — Thilo, Nov 08 '16 at 08:17
@delver there's just no need for the restriction: `if (a != null && a.intValue() == b.intValue())` handles all integer values. (You might also want to check that `b` isn't null). — Andy Turner, Nov 08 '16 at 08:27

Bathsheba · Accepted Answer · 2016-11-08T08:28:30.517

6

Because you have a -1 literal which is an int type, a will be auto-unboxed, (if and only if a == null is false else the expression is not evaluated) in the evaluation of the ternary conditional

a == null ? -1 : a;

Then since this expression is of type int, b is also auto-unboxed.

Therefore you can write if ((a == null ? -1 : a) == b), safe in the knowledge that Java will auto-unbox both a and b. So your explicit unboxing via the calls to intValue() is superfluous and your IDE is warning you of this. For the avoidance of doubt a is not auto-unboxed when the expression a == null is evaluated.

Personally I find this to be one of the most pernicious parts of Java. I'd seriously consider writing it the way you originally had it, and tell your IDE to suck it.

edited Nov 08 '16 at 08:28

answered Nov 08 '16 at 08:14

Bathsheba

231,907
34
361
483

What do you think about `if (!(a != null && a.intValue() == b.intValue()))`? – Boris Nov 08 '16 at 08:26
Personally I'd use `a.equals(b)`. – Bathsheba Nov 08 '16 at 08:27
2

Equivalent to, but more complex than `if (a == null || a.intValue() != b.intValue())` – Marko Topolnik Nov 08 '16 at 08:27
2

@delver in case you've not seen that transformation before, it's an application of [De Morgan's Laws](https://en.wikipedia.org/wiki/De_Morgan%27s_laws): `!(a && b) == !a || !b`. – Andy Turner Nov 08 '16 at 08:35

score -1 · Answer 2 · answered Nov 08 '16 at 08:15

-1

compareMayNull(Integer a, Integer b)

should do other than returning void...

maybe an int will be fine so you can do

(a == null ? -1 : a.compareTo(b)

answered Nov 08 '16 at 08:15

ΦXocę 웃 Пepeúpa ツ

47,427
17
69
97

Comparing two Integers 'Unnecessary unboxing' warning

2 Answers2