How to properly compare two Integers in Java?

Question

I know that if you compare a boxed primitive Integer with a constant such as:

Integer a = 4;
if (a < 5)

a will automaticall

Answer 1

== checks for reference equality, however when writing code like:

Integer a = 1;
Integer b = 1;

Java is smart enough to reuse the same immutable for a and b, so this is true: a == b. Curious, I wrote a small example to show where java stops optimizing in this way:

public class BoxingLol {
    public static void main(String[] args) {
        for (int i = 0; i < Integer.MAX_VALUE; i++) {
            Integer a = i;
            Integer b = i;
            if (a != b) {
                System.out.println("Done: " + i);
                System.exit(0);
            }
        }
        System.out.println("Done, all values equal");
    }
}

When I compile and run this (on my machine), I get:

Done: 128