Object obj = false ? new Long(1) : new Integer(1); System.out.println(obj.getClass());The smart once can guess that it is going to be "class java.lang.Long", otherwise there won't be a question. Now can you answer why? To be honest, I was a surprised by such an outcome, but apparently, according to JLS that is correct behaviour. Here is quotation from "15.25 Conditional Operator ? :":

*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.
- If one of the second and third operands is of type boolean and the type of the other is of type Boolean, then the type of the conditional expression is boolean.
- 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.
**Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:**- If one of the operands is of type byte or Byte and the other is of type short or Short, then the type of the conditional expression is short.
- If one of the operands is of type T where T is byte, short, or char, and the other operand is a constant expression of type int whose value is representable in type T, then the type of the conditional expression is T.
- If one of the operands is of type Byte and the other operand is a constant expression of type int whose value is representable in type byte, then the type of the conditional expression is byte.
- If one of the operands is of type Short and the other operand is a constant expression of type int whose value is representable in type short, then the type of the conditional expression is short.
- If one of the operands is of type; Character and the other operand is a constant expression of type int whose value is representable in type char, then the type of the conditional expression is char.
**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. Note that binary numeric promotion performs unboxing conversion (§5.1.8) and value set conversion (§5.1.13).**

- Otherwise, the second and third operands are of types S1 and S2 respectively. Let T1 be the type that results from applying boxing conversion to S1, and let T2 be the type that results from applying boxing conversion to S2. The type of the conditional expression is the result of applying capture conversion (§5.1.10) to lub(T1, T2) (§15.12.2.7).

In the other words, it says that if second and third operands are convertible to primitive numeric types, then the result type is based on numeric promotion of converted values. Here how it looks after applying these rules:

Object obj = Long.valueOf(false ? (new Long(1L)).longValue() : (new Integer(1)).intValue());My opinion, it all looks like a magic, and at the end, that's the price Java paid for autoboxing. Well world is not perfect and that's it's beauty, isn't it? :)

## 2 comments:

Hi Stas,

Nice blog! Is there an email address I can contact you in private?

For more fun try this:

Double v = null

Double v2 = 1 < 2 ? v : 0.0;

And try to guess the value of v2 (hint: it isn't null).

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