What's wrong with Integer, Java?

Question

    public class Test {
    public static void main(String[] args) {
        Integer i=555,j=555;
        System.out.println(i==j); //false
        Integer l=5,n=5;
        System.out.println(l==n); //true
    }
}

Why, Java? How is that even possible?

http://stackoverflow.com/questions/8427416/why-is-true-for-some-integer-objects?rq=1 - Duplicate, except the other way around. — Mat, Aug 29 '13 at 20:19
Same reason why you almost always never compare `Strings` with `==`. — Dennis Meng, Aug 29 '13 at 20:19
@arshajii I'm probably oversimplifying, but understanding one will help in understanding the other. — Dennis Meng, Aug 29 '13 at 20:21
As mentioned by others, you are comparing objects and also you need to understand internals of Java for integers. — Sachin Thapa, Aug 29 '13 at 20:22
This is a widely answered question, since Java 5 was released almost 10 years. — Peter Lawrey, Aug 29 '13 at 20:38

score 2 · Answer 1 · answered Aug 29 '13 at 20:21

You're comparing the references of two different Integer class instances with the same value, so you must use the equals method (as it must be to compare equality between objects):

Integer i=555,j=555;
System.out.println(i==j); //false
Integer i=555,j=555;
System.out.println(i.equals(j)); //true

But Integer has a pool of Integer object instances for int values between -128 and 127. So when you do

Integer l=5,n=5;
System.out.println(l==n); //true

You receive true since l and n points to the same object reference.

What's wrong with Integer, Java?

1 Answers1