CODEWITHSUNDEEP
javabit-manipulation
Don Wool
Don Wool

Reputation: 1167

decoding a byte in java

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?

Upvotes: 1

Views: 344

Answers (4)

Şeyma Yıldız
Şeyma Yıldız

Reputation: 1

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.

Upvotes: 0

ricosrealm
ricosrealm

Reputation: 1636

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.

Upvotes: 1

Viktor Latypov
Viktor Latypov

Reputation: 14467

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

Upvotes: 3

Eugene Retunsky
Eugene Retunsky

Reputation: 13139

This code will clear the bit 6. But if the bit 7 is set, it will set all bits from 8 to 31 (due to casting byte to int)

Upvotes: 2

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