Question

public class Test {
    public static final Double DEFAULT_DOUBLE = 12.0;
    public static final Long DEFAULT_LONG = 1L;

    public static Double convertToDouble(Object o) {
        return (o instanceof Number) ? ((Number) o).doubleValue()
                : DEFAULT_DOUBLE;
    }

    public static Long convertToLong(Object o) {
        return (o instanceof Number) ? ((Number) o).longValue() : DEFAULT_LONG;
    }

    public static void main(String[] args){
        System.out.println(convertToDouble(null) == DEFAULT_DOUBLE);
        System.out.println(convertToLong(null) == DEFAULT_LONG);
    }
}

Answer 1

EDIT

The ternary operator does some type conversions under the hood. In your case you are mixing primitives and wrapper types, in which case the wrapper types gets unboxed, then the result of the ternary operator is “re-boxed”:

If one of the second and third operands is of primitive type T, and the type of the other is the result of applying boxing conversion (§5.1.7) to T, then the type of the conditional expression is T.

---

So your code is essentially equivalent to (apart from the typo where longValue should be doubleValue):

public static void main(String[] args){
    Double d = 12.0;
    System.out.println(d == DEFAULT_DOUBLE);

    Long l = 1L;
    System.out.println(l == DEFAULT_LONG);
}

Long values can be cached on some JVMs and the == comparison can therefore return true. If you made all the comparisons with equals you would get true in both cases.

Note that if you use public static final Long DEFAULT_LONG = 128L; and try:

Long l = 128L;
System.out.println(l == DEFAULT_LONG);

It will probably print false, because Long values are generally cached between -128 and +127 only.

Note: The JLS requires that char, byte and int values between -127 and +128 be cached but does not say anything about long. So your code might actually print false twice on a different JVM.