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

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 2

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);