Please help me to get a clue from this program

Question

I'm sorry to bother you with such stupid questions. I have to admit I'm just making my first steps in programming.
So I have a program like this and don't understand how it works:
// Masking sign extension.
class HexByte {
static public void main(String args[]) {
char hex[] = {
'0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'a', 'b', 'c', 'd', 'e', 'f'
byte b = (byte) 0xf1;
System.out.println("b = 0x" + hex[(b >> 4) & 0x0f] + hex[b & 0x0f]);
Specifically, how could 0xf1 be represented in binary? As far as I know 11110001 contains negative value. Moreover 0xf1 in 10's base is 241, so how can I convert it to byte? Doin so, I'm gonna lose its value (since the byte range is from -128 to 127). And why do I need it? I even have no idea how the variable would look like in binary after converting to byte, not mentioning last part of this program.
I would really appreaciate your help.
Thank you for your time.

Answer 1

AntShay wrote:
Thank you so much for your help guys. Now I got just 2 more questions. First of all. I guess sign-extension works only with negative values. Is that correct? When sign-extension takes place and I shift any value to the right by any number of positions, does it fill in all the previous bits with *1s*? If you can, please, tell me how exactly this works.Java primitives are all signed, so sign extension is always at work for right bit shifts. However, if you were to look into the code, it has masked away the 4 most significant bits using bit-wise and operation. Alternatively, you may use the unsigned shift (+>>>+) operator.
And by the way, is there any method to convert the value of byte to a binary string? I only figured that it might work this way: Integer.toBinaryString(b&0xff), but is there Byte.toBinaryString method or something?It is always okay to up cast, as sign extension will convert correctly. If you don't like the long integer output, you may do a substring() on the output. It is also not hard to roll your own.