Question

I’m understanding how >>> works. To do it I have this program:

public class Main {
    public static void main(String[] args) 
    {
        short i = 130;
        byte b = (byte)i;
        String a = Integer.toBinaryString(256 + (int) b);
        System.out.println(Integer.toBinaryString(i));
        System.out.println(a.substring(a.length() -8));
        System.out.println(b);

        byte c = (byte) (b >>> 2);
        String x = Integer.toBinaryString(256 + (int) c);

        System.out.println(x.substring(x.length() -8));
        System.out.println(c);
    }
}

And I get this output:

10000010
10000010
-126
11100000
-32

To show as binary, I found here how to show a byte as a binary string.

Operator >>> will add zeros, but I get this:

-126
11100000
-63

Instead of:

-126
10100000
-32

It is adding a 1 instead of 0:

11100000
10100000

What am I doing wrong? Maybe I don’t understand anything.

Answer 1

The problem is that b >>> 2 is first promoting b to an int with value -126, i.e.

11111111 11111111 11111111 10000010

When you shift that right by 2 with zero-extension, you get:

00111111 11111111 11111111 11100000

When that’s then converted back to a byte, it just lops off the first three words, giving 11100000, which is what you’re seeing.

See section 15.19 of the JLS for more details about bit-shifting.