Using char as an unsigned 16 bit value in Java?

Question

I need an unsigned 8 bit integer in Java, and char seems to be the only thing close to that. Although it's double the size, it's unsigned which makes it practical for what I want to use it for (writing a basic emulator which requires unsigned bytes). The problem is that I've heard other programmers say that one shouldn't use char in that manner and should just use int or so. Is this true, and why so?

Answer 1

If you need an unsigned 8 bit integer then use byte. It's easy to make it unsigned in arithemtic operations (where actually sign matters) as byteValue & 0xFF

Answer 2

In Java:

long: [-2^63 , 2^63 - 1]

int: [-2^31 , 2^31 - 1]

short: [-2^15 , 2^15 - 1]

byte: [-2^7 , 2^7 - 1]

char: [0 , 2^16 - 1]

You want an unsigned 8 bit integer means you want a value between [0, 2^8 - 1]. It is clearly to choose short/int/long/char.

Although char can be treated as an unsigned integer, I think It's a bad coding style to use char for anything but characters.

For example,

public class Test {
public static void main(String[] args) {
    char a = 3;
    char b = 10;

    char c = (char) (a - b);
    System.out.println((int) c); // Prints 65529
    System.out.println((short) c); // Prints -7

    short d = -7;
    System.out.println((int) d); // Prints -7, Please notice the difference with char
}

}

It is better to use short/int/long with conversion.

Answer 3

It is totally reasonable to use byte to represent an unsigned 8 bit integer with some minor conversion, or Guava's UnsignedBytes do the conversion to you.