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

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
– khelwood, Commented Mar 14, 2019 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
– oggioniw, Commented Mar 14, 2019 at 16:15
If needed you can do x.intValue() == y.intValue(), but it is better to just use x.equals(y) — Andreas
– Andreas, Commented Mar 14, 2019 at 16:17
@oggioniw If either might be null, you need to null-check it (or use Objects.equals). Using .equals or .intValue() or (int) all risk raising a NullPointerException. — khelwood
– khelwood, Commented Mar 14, 2019 at 16:22

Karol Dowbecki · Accepted Answer · 2019-03-14 16:22:55Z

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, 2019 at 16:22

answered Mar 14, 2019 at 16:14

Karol Dowbecki

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3 Comments

oggioniw Over a year ago

what if I cast them to unboxed integers like this (int) x == (int) y?

Andreas Over a year ago

For full answer, mention use of Objects.equals as commented by khelwood as solution to null issue.

Antoniossss Over a year ago

Also cache range is configurable since Java7

gati sahu · Accepted Answer · 2019-03-14 16:25:18Z

0

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;
    }

answered Mar 14, 2019 at 16:25

gati sahu

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