Is it safe to compare two `Integer` values with `==` in Java?

Question

I have this Java code:

public class Foo {
    public static void main(String[] args) {
         Integer x = 5;
         Integer y = 5;
         System.out.println(x == y);
    }
}

Is it guaranteed to print true on the console? I mean, is it comparing the two boxed integers by value (which is what I need to do) or by reference identity?

Also, will it be any different if I cast them to unboxed integers like this

public class Foo {
    public static void main(String[] args) {
         Integer x = 5;
         Integer y = 5;
         System.out.println((int) x == (int) y);
    }
}

It is comparing them by reference identity. Integers in the range `-128` to `127` are cached, which is why Integer instances are sometimes the same reference. But you are better off using `equals`. — khelwood, Mar 14 '19 at 16:14
It will work but is highly discouraged. Compare literals with `==`. Compare objects with `equals()` — Arnaud Denoyelle, Mar 14 '19 at 16:14
the thing is that if I write `(int) x == (int) y` the IDE (I'm using intellijIdea) tells me that the cast is unnnecessary — oggioniw, Mar 14 '19 at 16:15
@khelwood They will be the same. Caching is documented behavior. — shmosel, Mar 14 '19 at 16:17
If needed you can do `x.intValue() == y.intValue()`, but it is better to just use `x.equals(y)` — Andreas, Mar 14 '19 at 16:17
@shmosel Yes, they will be the same in this specific case. In general two integers of the same value _might_ be the same but are not necessarily. — khelwood, Mar 14 '19 at 16:19
@Andreas what if one of them is `null` and the other is `0`? — oggioniw, Mar 14 '19 at 16:21
@oggioniw If either might be null, you need to null-check it (or use [`Objects.equals`](https://docs.oracle.com/javase/7/docs/api/java/util/Objects.html#equals(java.lang.Object,%20java.lang.Object))). Using `.equals` or `.intValue()` or `(int)` *all* risk raising a `NullPointerException`. — khelwood, Mar 14 '19 at 16:22
@khelwood +1 for suggesting usage of [Objects.equals](https://docs.oracle.com/javase/7/docs/api/java/util/Objects.html#equals(java.lang.Object,%20java.lang.Object)) — oggioniw, Mar 14 '19 at 16:26

Karol Dowbecki · Accepted Answer · 2019-03-14T16:22:55.337

15

No, it's not the right way to compare the Integer objects. You should use Integer.equals() or Integer.compareTo() method.

By default JVM will cache the Integer values from [-128, 127] range (see java.lang.Integer.IntegerCache.high property) but other values won't be cached:

Integer x = 5000;
Integer y = 5000;
System.out.println(x == y); // false

Unboxing to int or calling Integer.intValue() will create an int primitive that can be safely compared with == operator. However unboxing a null will result in NullPointerException.

edited Mar 14 '19 at 16:22

answered Mar 14 '19 at 16:14

Karol Dowbecki

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what if I cast them to unboxed integers like this `(int) x == (int) y`? – oggioniw Mar 14 '19 at 16:22
1

For full answer, mention use of `Objects.equals` as [commented by khelwood](https://stackoverflow.com/questions/55167297/is-it-safe-to-compare-two-integer-values-with-in-java?noredirect=1#comment97074301_55167297) as solution to null issue. – Andreas Mar 14 '19 at 16:27
1

Also cache range is configurable since Java7 – Antoniossss Mar 14 '19 at 16:35

score 0 · Answer 2 · answered Mar 14 '19 at 16:25

No it is not the right way to compare to object directly.You can convert to int value and compare.

Java implementation

Where value is of type int.

public boolean equals(Object obj) {
        if (obj instanceof Integer) {
            return value == ((Integer)obj).intValue();
        }
        return false;
    }

Is it safe to compare two `Integer` values with `==` in Java?

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