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

== will still test object equality. It is easy to be fooled, however:

Integer a = 10;
Integer b = 10;

System.out.println(a == b); //prints true

Integer c = new Integer(10);
Integer d = new Integer(10);

System.out.println(c == d); //prints false

Your examples with inequalities will work since they are not defined on Objects. However, with the == comparison, object equality will still be checked. In this case, when you initialize the objects from a boxed primitive, the same object is used (for both a and b). This is an okay optimization since the primitive box classes are immutable.