Bitwise Xor of bytes

[Ernie Mcracken]

This is puzzling me,

 byte one = 1;
byte two = 2;
byte three = one ^ two;

If I use the bitwise Xor of two bytes why do I get the compile time error

possible loss of precision
found : int
required: byte
byte three = one ^ two;

I guess this means that one ^ two returns an int, but I don't understand why.

Thanks

[Matthew Brown]

It's one of the gotchas of Java. All integer operators result in either an int or a long. So any operator acting on primitives "smaller than" int returns an int.

E.g.
short + short = int
byte * short = int
byte ^ byte = int

I'm not entirely sure why it was designed like that, but there'll probably be someone along to explain it .

The result is, you need to write:

byte three = (byte)(one ^ two);

[Ernie Mcracken]

That is rather strange.
Thank you!

[Wouter Oet]

And the operators += -= *= /= %= &= ^= |= <<= >>= >>>= have a build it cast.
So byte a = 5; a += 5; is valid.