Puzzle (A Big Delight in Every Byte)

Consider the follow example.

package demo;

public class App {

  public static void main( String[] args ) {
    for ( byte b = Byte.MIN_VALUE; b < Byte.MAX_VALUE; b++ ) {
      if ( b == 0x90 )
        System.out.println( "Joy!" );
    }
  }
}

How many times the above program prints Joy? Never?

This example was taken from PUZZLE 24: A BIG DELIGHT IN EVERY BYTE in Java™ Puzzlers: Traps, Pitfalls, and Corner Cases.

“Simply put, 0x90 is an int constant that is outside the range of byte values. This is counterintuitive because 0x90 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.”