Question

The following program prints respectively ‘false’ and ‘true’:

Number n = true ? new Long(1) : new Double(2.0);
System.out.println(n instanceof Long);
System.out.println(n instanceof Double);

So it will not be a Long but a Double. However, it works as intended on normal classes:
Having

class B {}
class D1 extends B {}
class D2 extends B {}

this will print ‘true’:

B b = true ? new D1() : new D2();
System.out.println(b instanceof D1);

meaning that it is not working the same like the above example.

I’m sure that there is something related to autoboxing, but is it really the way it should work? Why she uses boxing, when the Number class is a superclass of both Long and Double, so that expression could be evaluated to Number?

It’s really a pain, because when printing the n, it prints as a double value. (I know it is easy to workaround, but drove me crazy)

Answer 1

Let’s take out the language lawyer’s book here: JLS §15.25

The type of a conditional expression is determined as follows:

• If the second and third operands have the same type (which may be the null type), then that is the type of the conditional expression.

Long and Double are not the same type – does not apply.

• 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.

Neither value is primitive – does not apply.

• If one of the second and third operands is of the null type and the type of the other is a reference type, then the type of the
  conditional expression is that reference type.

Neither value is null – does not apply.

• Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:
  • [… special cases for byte/short/char and their boxed equivalents…]
  • Otherwise, binary numeric promotion (§5.6.2) is applied to the operand types, and the type of the conditional expression is the promoted type of the second and third operands.

This rule does apply here, which means that the result type of the conditional operator is as if both values were unboxed. Is suppose the reasoning behind that was that otherwise Number n = bool ? 1 : 2.0 and Number n = bool ? new Long(1) : new Double(2.0) have different values. This behaviour would also be unexpected and – worse – inconsistent.