Bitwise & with 0xFF

Question

Folks, Looking at a piece of code, didn't understand why do we need to do a bitwise & with 0xFF.
StringBuffer sb = new StringBuffer("F1");
short hexNum = Short.parseShort(sb.toString(), 16);
b[bufIdx] = (byte) (*hexNum & 0xFF*);
As we are casting it to byte, the upper byte would be truncated anyways? If I don't get it right, can someone please explain the logic?
TIA!

Answer 1

From the JLS:
A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T. In addition to a possible loss of information about the magnitude of the numeric value, this may cause the sign of the resulting value to differ from the sign of the input value.
--[http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#25363]
So yeah, I personally can't see any case in which this:
byte b = (byte)(someInt);would be any different than this:
byte b = (byte)(someInt & 0xFF);Since the JLS explicitly states that all but the last 8 bits will be discarded, and ( & 0xFF ) is guaranteed not to change anything in the last 8 bits.
If you had an example in which the two results would be different, I might be able to explain why...