Integer wrapper class and == operator - where is b

Question

This question already has answers here:

Integer wrapper objects share the same instances only within the value 127? [duplicate] (5 answers)

Closed 5 years ago.

Integer integer1 = 127;
Integer integer2 = 127;
System.out.println(integer1 == integer2);//true

integer1 = 128;
integer2 = 128;
System.out.println(integer1 == integer2);//false

I found it returns == (if it is) under the range of -128 - 127 , why is there such specification ?

Answer 1

== will return true if it's the exact same object. Boxing integers in Java 'intern' numbers within that that range, so any boxed version of such a number will result in the exact same object.

To get avoid this effect in comparisons, use .equals()

System.out.println(integer1.equals(integer2));

Answer 2

Because of this code in Integer.valueOf(int):

public static Integer valueOf(int i) {
    if(i >= -128 && i <= IntegerCache.high)
        return IntegerCache.cache[i + 128];
    else
        return new Integer(i);
}

Explanation:

Integer integer1 = 127 is a shortcut for Integer integer1 = Integer.valueOf(127), and for values between -128 and 127 (inclusive), the Integers are put in a cache and returned multiple times, while higher and lower numbers generate new Integers each time.