Question

I’m wondering why there is an inconsistency with the code below. I would expect the same output, but when using the inline conditional statement, it appends a .0 to the string.
Do I have some error in my code?

    double d = 10.1;

    String rounded = (false ? d : Math.round(d)) + "";
    System.out.println(rounded);//10.0

    rounded = Math.round(d) + "";
    System.out.println(rounded);//10

Answer 1

Math.round returns a long, therefore the two operands of the conditional operator do not have the same type, and thus a more complex rule is followed to determine the type of the overall operation, as defined in JLS §15.25:

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).

And from 5.6.2, binary numeric promotion:

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

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And to illustrate the pitfalls with the conditional operator and for some fun, from Java puzzlers (puzzle 8):

char x = 'X';
int i = 0;
System.out.print(true ? x : 0); // prints X
System.out.print(false ? i : x); // prints 88 => (int)X 

Also, check out the Hamlet and Elvis examples (video links).