Java, store in byte higher value than 127

Question

For an assignment I get some bytes, make some calculations on their Binary values and have to return that calculated Stuff in a Byte Array.

My Problem is now, that byte only stores up to 127, but my values can be up to 2^8-1 (11111111). I already tried to convert it to hex, but it´s of course not working either.

So, is that possible, if yes how, or is that exercise not possible like that?

Answer 1

If I'm understanding your question, before you perform any calculations you need to AND (&) your bytes against 255, so you'll be dealing with values from 0 to 255 instead of -128 to 127. When you assign a value higher than 127 to a byte it overflows, so 128 as a byte in java is -128, but after you AND (&) -128 against 255 you'll get +128.

// This prints 0 to 127, then -128 to -1
public static void main(String[] args) {
    for (int i = 0; i <= 255; i++) {
        System.out.println(((byte)i));
    }
}

// Where this will print 0 to 255
public static void main(String[] args) {
    for (int i = 0; i <= 255; i++) {
        System.out.println(((byte)i) & 255);
    }
}

Answer 2

Typically, in Java, this number would be represented as an int . That gives you more than enough room to store your values. If you need to represent it as a byte, for transmitting it somewhere or something like that, then you would cast it like

int i = 255;
byte b = (byte) (i & 0xff);

but in this case, it will still be negative, because byte is signed in Java.

Also, hexadecimal literals are by default of type int , as you've probably found out. So 0xff does have the value of 255 like you intend, but it's an int , not a byte .