When using == for a primitive and a boxed value, is autoboxing done, or is unboxing done

Question

The following code compiles (with Java 8):

Integer i1 = 1000; int i2 = 1000; boolean compared = (i1 == i2); 

But what does it do?

Unbox i1:

boolean compared = (i1.intvalue() == i2); 

or box i2:

boolean compared = (i1 == new Integer(i2)); 

So does it compare two Integer objects (by reference) or two int variables by value?

Note that for some numbers the reference comparison will yield the correct result because the Integer class maintains an internal cache of values between -128 to 127 (see also the comment by TheLostMind). This is why I used 1000 in my example and why I specifically ask about the unboxing/boxing and not about the result of the comparison.

Answer 1

It is defined in the JLS #15.21.1:

If the operands of an equality operator are both of numeric type, or one is of numeric type and the other is convertible (§5.1.8) to numeric type, binary numeric promotion is performed on the operands (§5.6.2).

And JLS #5.6.2:

When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order:

• If any operand is of a reference type, it is subjected to unboxing conversion [...]

So to answer your question, the Integer is unboxed into an int.