Why Does Java's Integer Boxing Produce Different Results for `Integer == Integer` Comparisons?

Weird Java Integer Boxing

Java's integer boxing mechanism can leave you slightly confused, especially when comparing reference equality.

Problem

Consider the following code snippet:

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

        Integer c = 100, d = 100;
        System.out.println(c == d); // true
    }
}

When run, it will generate:

false
true

First line The result is expected because a and b are different objects. But what is the result of the second line?

Answer

The true result of the second line is guaranteed by the language specification. According to Section 5.1.7:

If the value p to be boxed is true, false, byte or char in the range u0000 to u007f, or int or short between -128 and 127, then The results of any two boxing transformations on p, r1 and r2, are always equal.

While the second line of output is guaranteed, the first line is not (see the last paragraph quoted below):

Ideally, given a primitive value p Boxing always generates the same reference. In practice, this may not be possible using existing implementation techniques. The above rule is a pragmatic compromise. The last clause above requires that certain common values ​​be always boxed into indistinguishable objects. Implementations MAY cache it, either lazily or immediately.

For other values, this representation does not allow the programmer to make any assumptions about the identity of the boxed value. This allows (but does not force) sharing of some or all of these references.

This ensures that in most common cases the behavior will be the desired behavior without unnecessary loss of performance, especially on small devices. Less memory-constrained implementations can cache all characters and shorts, as well as integers and longs in the range -32K to 32K.

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