Weird Integer boxing in Java

Question

I just saw code similar to this:

public class Scratch
{
    public static void main(String[] args)
    {
        Integer a = 1000, b = 1000;
        System.o         


        

Answer 1

Integer Cache is a feature that was introduced in Java Version 5 basically for :

1. Saving of Memory space
2. Improvement in performance.

Integer number1 = 127;
Integer number2 = 127;

System.out.println("number1 == number2" + (number1 == number2); 

OUTPUT: True

---

Integer number1 = 128;
Integer number2 = 128;

System.out.println("number1 == number2" + (number1 == number2);

OUTPUT: False

HOW?

Actually when we assign value to an Integer object, it does auto promotion behind the hood.

Integer object = 100;

is actually calling Integer.valueOf() function

Integer object = Integer.valueOf(100);

Nitty-gritty details of valueOf(int)

    public static Integer valueOf(int i) {
        if (i >= IntegerCache.low && i <= IntegerCache.high)
            return IntegerCache.cache[i + (-IntegerCache.low)];
        return new Integer(i);
    }

Description:

This method will always cache values in the range -128 to 127, inclusive, and may cache other values outside of this range.

When a value within range of -128 to 127 is required it returns a constant memory location every time. However, when we need a value thats greater than 127

return new Integer(i);

returns a new reference every time we initiate an object.

== operators in Java compares two memory references and not values.

Object1 located at say 1000 and contains value 6.
Object2 located at say 1020 and contains value 6.

Object1 == Object2 is False as they have different memory locations though contains same values.