Expressive Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 14

of long , or forgetting to add the first operand. What's going on here?
Decimal literals have a nice property that is not shared by hexadecimal or octal literals: Decimal
literals are all positive [JLS 3.10.1]. To write a negative decimal constant, you use the unary
negation operator ( - ) in combination with a decimal literal. In this way, you can write any int or
long value, whether positive or negative, in decimal form, and negative decimal constants are
clearly identifiable by the presence of a minus sign. Not so for hexadecimal and octal literals.
They can take on both positive and negative values. Hex and octal literals are negative if their
high-order bit is set. In this program, the number 0xcafebabe is an int constant with its high-order
bit set, so it is negative. It is equivalent to the decimal value -889275714 .
The addition performed by the program is a mixed-type computation : The left operand is of type
long , and the right operand is of type int . To perform the computation, Java promotes the int
value to a long with a widening primitive conversion [JLS 5.1.2] and adds the two long values.
Because int is a signed integral type, the conversion performs sign extension : It promotes the
negative int value to a numerically equal long value.
The right operand of the addition, 0xcafebabe , is promoted to the long value
0xffffffffcafebabeL . This value is then added to the left operand, which is 0x100000000L . When
viewed as an int , the high-order 32 bits of the sign-extended right operand are -1 , and the high-
order 32 bits of the left operand are 1 . Add these two values together and you get 0 , which explains
the absence of the leading 1 digit in the program's output. Here is how the addition looks when done
in longhand. (The digits at the top of the addition are carries.)
1111111
0xffffffffcafebabeL
+ 0x0000000100000000L
0x00000000cafebabeL
Fixing the problem is as simple as using a long hex literal to represent the right operand. This
avoids the damaging sign extension, and the program prints the expected result of 1cafebabe :
public class JoyOfHex {
public static void main(String[] args) {
System.out.println(
Long.toHexString(0x100000000L + 0xcafebabe L ));
}