Why does the xor operator on two bytes produce an int?

Question

        //key & hash are both byte[]
        int leftPos = 0, rightPos = 31;
        while(leftPos < 16) {
            //possible loss of precision. required: byte, found: int
            key[leftPos] = hash[leftPos] ^ hash[rightPos];
            leftPos++;
            rightPos--;
        }

Why would a bitwise operation on two bytes in Java return an int? I know I could just cast it back to byte, but it seems silly.

Answer 1

every arithmetic operation on bytes promotes them to int.

Answer 2

There is no Java bitwise operations on two bytes. Your code implicitly and silently converts those bytes to a larger integer type (int), and the result is of that type as well.

You may now question the sanity of leaving bitwise operations on bytes undefined.

Answer 3

Because the language spec says so. It gives no reason, but I suspect that these are the most likely intentions:

• To have a small and simple set of rules to cover arithmetic operations involving all possible combinations of types
• To allow an efficient implementation - 32 bit integers are what CPUs use internally, and everything else requires conversions, explicit or implicit.

Answer 4

This was somewhere down in the answers to one of the similar questions that people have already pointed out:

http://blogs.msdn.com/oldnewthing/archive/2004/03/10/87247.aspx

Answer 5

If it's correct and there are no value that can cause this loss of precision, in other words : "impossible loss of precision" the compiler should shut up ... and need to be corrected, and no cast should be added in this :

byte a = (byte) 0xDE; byte b = (byte) 0xAD; byte r = (byte) ( a ^ b);

-- http://rzr.online.fr/q/java