Question

Hey I am trying to understand the following code snippet.

public static void main(String[] args) {
    Integer i1 = 1000;
    Integer i2 = 1000;
    if(i1 != i2) System.out.println("different objects");
    if(i1.equals(i2)) System.out.println("meaningfully equal");
    Integer i3 = 10;
    Integer i4 = 10;
    if(i3 == i4) System.out.println("same object");
    if(i3.equals(i4)) System.out.println("meaningfully equal");
}

This method runs all of the println instructions. That is i1 != i2 is true, but i3 == i4. At first glance this strikes me as strange, they should be all different as references. I can figure out that if I pass the same byte value (-128 to 127) to i3 and i4 they will be always equal as references, but any other value will yield them as different.

I can’t explain this, can you point me to some documentation or give some helpful insights?

Thank you

Answer 1

Autoboxing int values to Integer objects will use a cache for common values (as you’ve identified them). This is specified in the JLS at §5.1.7 Boxing Conversion:

If the value p being boxed is true, false, a byte, a char in the range \u0000 to \u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 == r2.

Note that this will only be applied when the language auto-boxes a value for you or when you use Integer.valueOf(). Using new Integer(int) will always produce a new Integer object.

Minor hint: a JVM implementation is free to cache values outside of those ranges as well, because the opposite is not specified. I’ve not yet seen such an implementation, however.