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
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
orByte
and the other is of typeshort
orShort
, then the type of the conditional expression isshort
.[...]
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
.
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