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

We should always go for equals() method for comparison for two integers.Its the recommended practice.

If we compare two integers using == that would work for certain range of integer values (Integer from -128 to 127) due to JVM's internal optimisation.

Please see examples:

Case 1:

Integer a = 100; Integer b = 100;

if (a == b) {
    System.out.println("a and b are equal");
} else {
   System.out.println("a and b are not equal");
}

In above case JVM uses value of a and b from cached pool and return the same object instance(therefore memory address) of integer object and we get both are equal.Its an optimisation JVM does for certain range values.

Case 2: In this case, a and b are not equal because it does not come with the range from -128 to 127.

Integer a = 220; Integer b = 220;

if (a == b) {
    System.out.println("a and b are equal");
} else {
   System.out.println("a and b are not equal");
}

Proper way:

Integer a = 200;             
Integer b = 200;  
System.out.println("a == b? " + a.equals(b)); // true

I hope this helps.