Java unsigned number processing

Java does not have unsigned numbers, which can cause a lot of problems.

static byte[] x = {(byte) 0xff,(byte) 0xff};
    
    public static void main(String[] args) throws IOException {
    
        byte a = x[0];
        
        int z = a&0xff;
        System.out.println(z);
        
    }

int z = a&0xff,

First of all, a is of byte type. When performing & operation on it, it will be converted to int type first, that is,

　1111 1111 1111 1111 1111 1111 1111 1111

&

　0000 0000 0000 0000 0000 0000 1111 1111

=0000 0000 0000 0000 0000 0000 1111 1111

The result is 255. If the &0xff operation is not performed and assigned directly to z, the obtained value is -1.