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);
    }
}

Answer 1

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.