Question

The code that I can’t understand does this :

int decodeTimeStampByte(final byte timeByte) {
   return timeByte & (~64);
}

So for instance, if I get the byte 4c (which is ASCII L), what exactly would the above function do to it? How about the byte 44?

Answer 1

The ‘~’ is bitwise ‘not’, so 64 = 0x40 = 0100000b and ~64 = 1011111b (the lower 5 bits set).

Then ‘&’ is bitwise ‘and’ and it leaves just the 5 lower bits of timeByte. So, basically, it is a truncation of timeByte to 0..63 range.

decodeTimeStampByte(4c) = 0xC (12)

decodeTimeStampByte(44) = 44

P.S. Yes, I forgot the higher bits. ~64 = 1011111b.

It is either a bug in the code or some intention to leave the sign bit (the 7-th bit) in place.

P.P.S. Seems like an ancient bit-hack to squeeze some more performance