Help needed in understanding conversion alghorithm from byte to hex

Question

Hi, I'm studying the following code:
public static char[] byteToHex(byte[] data) {
  char[] retValue = new char[data.length * 2];
  int value = 0;
  int highIndex = 0;
  int lowIndex = 0;
  for (int i = 0; i < retValue.length; i++) {
    value = (data[i] + 256) % 256;
    highIndex = value >> 4;
    lowIndex = value & 0x0f;
    retValue[i * 2 + 0] = hexTable[highIndex];
    retValue[i * 2 + 1] = hexTable[lowIndex];
  return retValue;
}There are few things (the most important) which I don't understand about the above code.
I understood that what's returned has got double size related to what's passed in because a char takes 16 bits while a byte takes 8.
1) I don't understand why each byte must be first added 256 and then % with 256 (returning the same value - Is this to eliminate negative values?)
2) I do understand that each byte is transformed in two hexadecimal values: one is the highIndex (first 8 bits) and the second is the lowerIndex (last 8 bits) and that each value is tranformed in its hexadecimal value from the array of hex values.
3) What I don't really understand is why the highIndex is calculated as: value >> 4
and the lowest index is calculated as value 0x0f (is this last also to eliminate negative values?)
If someone could clarify this for me, I'd be very grateful.
Thanks.
Marco

Answer 1

So, does this mean that we add 256 to eliminate the sign?No. You need the whole line to convert a signed byte into an int between 0 and 255.
A simpler way to do this would be
value = data[i] & 0xFF;
This moves down the higher bits so that it turnsinto lower bits. i.e. we need it to >be between 0 and
15.
Is this shifted of 4 because Math.pow(2, 4) = 16.0?Doing in this case, x >>4 is the same as x / 16
This leaves only the lowest 4 bits.Is the following what happens?There are no char values produced. Using 0000 as an example is not a good idea as you can change it in many ways and it is still 0000
>
Received as initial value:
byte: 0000 0000
What we need to obtain:
char: 0000 0000 0000 0000
The first 4 bits of the above byte are shifted of 4
positions to find the hexadecimal equivalent (if from
2 I want to get to 16 I have to do the opposite of
powering a number by 4); The last four bits of the
byte are extracted because of the '&' operator with
0x0f (which in binary is 1111 - Therefore all the '1'
are kept)?Yes.