用Java解码一个字节

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

PS 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.

PPS Seems like an ancient bit-hack to squeeze some more performance

Answer 2

该代码将清除第6位。但是，如果设置了第7位，它将把所有位设置为8到31（由于将字节转换为int）

Answer 3

This function is returning the lower 6 bits for positive values and clearing the 7th bit for negative values. So, 2^6=64, 64 = 1000000 in binary, ~64 = 0111111 in binary would mask values between [0..63] and [-128..-65] of timeByte.

Answer 4

This function is returning the lower 6 bits for positive values and clearing the 7th bit for negative values. So, 2^6=64, 64 = 1000000 in binary, ~64 = 0111111 in binary would mask values between [0..63] and [-128..-65] of timeByte .