Expressive Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 16

< Day Day Up >
Puzzle 6: Multicast
Casts are used to convert a value from one type to another. This program uses three casts in
succession. What does it print?
public class Multicast {
public static void main(String[] args) {
System.out.println((int) (char) (byte) -1);
}
}
Solution 6: Multicast
This program is confusing any way you slice it. It starts with the int value -1 , then casts the int to
a byte , then to a char , and finally back to an int . The first cast narrows the value from 32 bits
down to 8, the second widens it from 8 bits to 16, and the final cast widens it from 16 bits back to
32. Does the value end up back where it started? If you ran the program, you found that it does not.
It prints 65535 , but why?
The program's behavior depends critically on the sign extension behavior of casts. Java uses two's-
complement binary arithmetic, so the int value -1 has all 32 bits set. The cast from int to byte is
straightforward. It performs a narrowing primitive conversion [JLS 5.1.3], which simply lops off all
but the low-order 8 bits. This leaves a byte value with all 8 bits set, which (still) represents -1.
The cast from byte to char is trickier because byte is a signed type and char unsigned. It is usually
possible to convert from one integral type to a wider one while preserving numerical value, but it is
impossible to represent a negative byte value as a char . Therefore, the conversion from byte to
char is not considered a widening primitive conversion [JLS 5.1.2], but a widening and narrowing
primitive conversion [JLS 5.1.4]: The byte is converted to an int and the int to a char .
All of this may sound a bit complicated. Luckily, there is a simple rule that describes the sign
extension behavior when converting from narrower integral types to wider: Sign extension is
performed if the type of the original value is signed; zero extension if it is a char , regardless of
the type to which it is being converted . Knowing this rule makes it easy to solve the puzzle.
Because byte is a signed type, sign extension occurs when converting the byte value -1 to a char .
The resulting char value has all 16 bits set, so it is equal to 2 16 - 1, or 65,535. The cast from char
to int is also a widening primitive conversion, so the rule tells us that zero extension is performed