Strange java behaviour with conditional operator. Is it a bug?

Question

Can you please run the below and explain?

Object o = true ? new Integer(1) : new Double(2.0);
System.out.println(o);

I found that surprising as someone would expect 1 to be printed and not 1.0

Answer 1

It's not a surprise at all, although it might seem like one. The behaviour is specified in JLS §15.25 - Conditional Operator :

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 .

  [...]

• 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 value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).}

So the Integer and Double types are unboxed to their respective primitive counterparts - int and double , as a process of binary numeric promotion. And then the type of the conditional operator is the promoted type of int and double , which is double . Hence the result is 1.0 . And of course the final result is then boxed back to Double .

Answer 2

Here is an article published in DZone yesterday talking about it:

Java auto unboxing gotcha

Funny enough, the example code looks similar...