Loopy Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 64

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Puzzle 24: A Big Delight in Every Byte
This program loops through the byte values, looking for a certain value. What does the program
print?
public class BigDelight {
public static void main(String[] args) {
for (byte b = Byte.MIN_VALUE; b < Byte.MAX_VALUE; b++) {
if (b == 0x90)
System.out.print("Joy!");
}
}
}
Solution 24: A Big Delight in Every Byte
The loop iterates over all the byte values except Byte.MAX_VALUE , looking for Ox90 . This value fits
in a byte and is not equal to Byte.MAX_VALUE , so you might think that the loop would hit it once and
print Joy! on that iteration. Looks can be deceiving. If you ran the program, you found that it prints
nothing. What happened?
Simply put, Ox90 is an int constant that is outside the range of byte values. This is counterintuitive
because Ox90 is a two-digit hexadecimal literal. Each hex digit takes up 4 bits, so the entire value
takes up 8 bits, or 1 byte. The problem is that byte is a signed type. The constant 0x90 is a positive
int value of 8 bits with the highest bit set. Legal byte values range from -128 to +127, but the int
constant 0x90 is equal to +144.
The comparison of a byte to an int is a mixed-type comparison . If you think of byte values as
apples and int values as oranges, the program is comparing apples to oranges. Consider the
expression ((byte)0x90 == 0x90) . Appearances notwithstanding, it evaluates to false . To
compare the byte value (byte)0x90 to the int value 0x90 , Java promotes the byte to an int with a
widening primitive conversion [JLS 5.1.2] and compares the two int values. Because byte is a
signed type, the conversion performs sign extension , promoting negative byte values to numerically
equal int values. In this case, the conversion promotes (byte)0x90 to the int value -112 , which is
unequal to the int value 0x90 , or +144 .