Question

Java has 2 bitshift operators for right shifts:

>> shifts right, and is dependant on the sign bit for the sign of the result

>>> shifts right and shifts a zero into leftmost bits

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/op3.html

This seems fairly simple, so can anyone explain to me why this code, when given a value of -128 for bar, produces a value of -2 for foo:

byte foo = (byte)((bar & ((byte)-64)) >>> 6);

What this is meant to do is take an 8bit byte, mask of the leftmost 2 bits, and shift them into the rightmost 2 bits. Ie:

initial = 0b10000000 (-128)
-64 = 0b11000000
initial & -64 = 0b10000000
0b10000000 >>> 6 = 0b00000010

The result actually is -2, which is

0b11111110

Ie. 1s rather than zeros are shifted into left positions

Answer 1

It’s because the & is actually performing promotion to int – which leaves an awful lot of “1” bits. You’re then shifting right, leaving the leftmost 2 bits as 0, but then ignoring those leftmost bits by casting back to byte.

This becomes clearer when you separate out the operations:

public class Test
{
    public static void main(String[] args)
    {
        byte bar = -128;
        int tmp = (bar & ((byte)-64)) >>> 6;
        byte foo = (byte)tmp;
        System.out.println(tmp);
        System.out.println(foo);
    }
}

prints

67108862
-2

So to do your bit arithmetic again:

initial = 0b10000000 (-128)
-64 = 0b11000000
initial & -64 = 0b11111111111111111111111110000000 // it's an int now
0b10000000 >>> 6 = 0b00111111111111111111111111100000 // note zero-padding
(byte) (0b10000000 >>> 6) = 11100000 // -2

Even if you get the right result out of the & operation (by casting at that point), >>> will promote the first operand to int first anyway.

EDIT: The solution is to change how you mask things. Instead of masking by -64, mask by just 128+64=192=0xc0 instead:

byte foo = (byte)((bar & 0xc0) >>> 6);

That way you really only get left with the two bits you want, instead of having a load of 1s in the most significant 24 bits.