Bytes in Java with hex

Question

buffer is a bytebuffer .I'm getting a lost of percision error with this.

    byte myPort = buffer.get(0); // Might need to change this depending on byte order
    switch(myPort){
        case 0xF1: // Chat service
            break;
        case 0xF2: // Voice service
            break;
        case 0xF3: // Video service
            break;
        case 0xF4: // File transfer service
            break;
        case 0xF5: // Remote login
            break;
    }

Apparently, 0xFF is not a byte in java and its really confusing me. I don't know if I'm losing it but isn't 0xF a nibble and 0xFF a byte? Apparently my ide netbeans allows byte values all the way up to 127. This seems to be a problem with signed values, but I'm not sure why.

Thanks for any help.

Answer 1

As already stated in the comments, byte is [-128 .. 127].

You should convert your byte from the byte buffer to int, to preserve the "unsigned byte" range you assume (and which does not exist in Java):

  int myPort = buffer.get(0) & 0xff; // Might need to change this depending on byte order
  switch(myPort){
      case 0xF1: // Chat service
          break;
      case 0xF2: // Voice service
          break;
      case 0xF3: // Video service
          break;
      case 0xF4: // File transfer service
          break;
      case 0xF5: // Remote login
          break;
  }

Note that simply assigning the byte to an int would not work due to sign extension:

  byte val = (byte) 0xb6;  
  int myPort = val;

results in myPort = -74 (0xffffffb6), while

  byte val = (byte) 0xb6;
  int myPort = val & 0xff;

results in myPort = 182 (0xb6).

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See Why doesn't Java support unsigned ints? for some good background information on the absence of unsigned data types in Java