Weird behaviour of ternary conditional with boxed types in Java [duplicate]

Question

I had this piece of code in my application (simplified version):

Object result; if (check)     result = new Integer(1); else     result = new Double(1.0); System.out.println(result); return result; 

Then I decided to refactor the if-else statement to a ternary conditional expression so my code is more concise:

Object result = check ? new Integer(1) : new Double(1.0); System.out.println(result); return result; 

It turned out that in case check is true the two versions print different results:

1 

or:

1.0 

Isn't the ternary conditional equivalent to the corresponding if-else?

Answer 1

The if/else and the conditional (ternary) expression are not quite equivalent. The result of the conditional expression must have a type.

You're observing the effects of numeric type promotion (or type coercion).

Here's an excerpt from the language spec (see here), from the section describing the return value of the conditional expression:

Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:

The final such case is (I've omitted the other cases here):

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.

Here's the additional spec related to binary numeric promotion:

Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

• If either operand is of type double, the other is converted to double.

It's the first case (following cases omitted). double always wins.

So regardless of the order of the 2nd and 3rd operands in your conditional expression, the return type of the expression will be promoted to double.