Question

I’m a bit confused regarding a conversion from bytes to integers. Consider the following code:

byte[] data = new byte[] { 0, (byte) 0xF0 };

int masked = data[0] << 8 & 0xFF | data[1] & 0xFF; //240
int notMasked = data[0] << 8 | data[1]; //-16

Because bytes in java are signed, data[1] is not 240 decimal, but rather the 2’s complement, -16. However, it should still be, in binary: 0x11110000 so, why do I need to do data[1] & 0xFF ?

Is Java converting everything to Integer before passing it to the | operator? Why does &0xFF make a difference then?

Answer 1

Java bytes are signed (unfortunately) – so when you promote the value to an int in order to perform the bitwise |, it ends up being sign-extended as 0xFFFFFFF0. That then messes up the | with data[0]. The masking with & 0xff converts it to an integer value of 240 (just 0x000000F0) instead.

However, you’ve stlil got a problem. This code:

int masked = data[0] << 8 & 0xFF | data[1] & 0xFF;

should be:

int masked = ((data[0] & 0xff) << 8) | (data[1] & 0xFF);

… otherwise you’re masking after the shift, which won’t work. I’ve added brackets because I’m never sure of the predence of &, << and |…