Unsigned Bytes in Java

Question

Bytes in Java are signed by default. I see on other posts that a workaround to have unsigned bytes is something similar to that: int num = (int) bite & 0xFF

Could someone please explain to me why this works and converts a signed byte to an unsigned byte and then its respective integer? ANDing a byte with 11111111 results in the same byte - right?

Answer 1

A typecast has a higher precedence than the & operator. Therefore you're first casting to an int, then ANDing in order to mask out all the high-order bits that are set, including the "sign bit" of the two's complement notation which java uses, leaving you with just the positive value of the original byte. E.g.:

let byte x = 11111111 = -1
then (int) x = 11111111 11111111 11111111 11111111
and x & 0xFF = 00000000 00000000 00000000 11111111 = 255

and you've effectively removed the sign from the original byte.

Answer 2

ANDing a byte with 11111111 results in the same byte - right?

Except you're ANDing with 00000000000000000000000011111111, because 0xFF is an int literal - there are no byte literals in Java. So what happens is that the byte is promoted to int (the typecast is unnecessary), has its sign extended (i.e. keeps the possibly negative value of the byte, but then the sign extension is reverted by ANDing it with all those zeroes. The result is an int that has as its least significant bits exactly the former byte and thus the value the byte would have had were it unsigned.