Comparing Long object type with primitive int using ==

Question

I have a method that returns a Long object datatype via invocation of: resp.getResultCode(). I want to compare it HttpStatus.GONE.value() which actually just returns a primitive int value of 410. Would the Long unbox itself to properly compare with the int primitive?

if(resp.getResultCode() == HttpStatus.GONE.value()){
  // code inside..
}

Answer 1

Here's the JLS explanation

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

If the promoted type of the operands is int or long, then an integer equality test is performed.

So the Long is unboxed to long. And numeric promotion is applied to int to make it a long. Then they are compared.

Consider that case where long would be "demoted" to an int, you'd have cases like this

public static void main(String[] args) throws Exception {
    long lvalue = 1234567891011L;
    int ivalue = 1912277059;
    System.out.println(lvalue == ivalue); // false
    System.out.println((int) lvalue == ivalue); // true, but shouldn't
}