Bitwise Shift Clarification / Bug ?

Question

Java Tutorial: Shift & Logical Operators...
has no example to clarify difference between op1 >> op2 vs op1 >>> op2 (only says latter is unsigned).
To help understand, I wrote following code...
public class BitwiseDemo {
public static void main(String[] args) {
int op1 = -13;
int op2 = 2;
int res1 = ( op1 >> op2 );
int res2 = ( op1 >>> op2 );
System.out.println("op1 >> op2 " + res1 );
System.out.println("op1 >>> op2 " + res2 );
I got result...
op1 >> op2 -4
op1 >>> op2 1073741820 .... (big)
In 2's complement arithmetic which Java uses...
-13 =1111 1111 1111 1111 1111 1111 1111 0011
If we shift -13 to right by two retaining sign bit
AND padding with 1 (not 0)...
-4 =1111 1111 1111 1111 1111 1111 1111 1100
If we shift -13 to right by two including sign bit
AND padding with 0 (as doco says)...
(big)=0011 1111 1111 1111 1111 1111 1111 1100
I suggest doco should read...
pad with leading sign bit
pad with zero
Q:Is the above correct ?
Q:Is there a part of Java doco which gives an example
of this? I don't think it makes a difference for positive
numbers.
Q:Do I get some award for this great piece of learning?
Instead of using class int, I use class byte but does not compile...
BitwiseDemo.java:10 (&11): possible loss of precision
found : int
required : byte
byte res1 = ( op1 >> op2 );
^
I expected it to compile and have the answers...
In 2's complement arithmetic which Java uses...
-13 =1111 0011
If we shift -13 to right by two retaining sign bit
AND padding with 1...
-4 =1111 1100 (>>)
If we shift -13 to right by two including sign bit...
60 =0011 1100 (>>>)
I also get no compile with class short.
Instead of using class int, I use class long...
I get -4 and 46116860184273877900
Q: Does this mean that the >> and >>> operators use class int as a minimum? Is this in doco?
Rgds
Tony

Answer 1

The program I used was...
public class BitDemo {
public static void main(String[] args) {
byte b= (byte)0x80;
byte c= (byte)0x80;
b >>>= 1;
c >>= 1;
System.out.println("b >>>= 1 " + b );
System.out.println("b >>= 1 " + c );
This produced -64 in both cases as you said.
PS I used (byte) to type cast the 0x80 assignment.
I also used byte b= -128; which did the same thing.
My logic says...
0x80 =1000 0000
= =1100 0000 (-64)
= =0100 0000 (+64)
So my maths doesn't agree ! The sign is wrong.
Then I tried the program with int instead of byte...
public class BitDemo {
public static void main(String[] args) {
int b= 0x80000000;
int c= 0x80000000;
b >>>= 1;
c >>= 1;
System.out.println("b >>>= 1 " + b );
System.out.println("b >>= 1 " + c );
Result...
b >>>=1 1073741824
b >>=1 -1073741824
So using int I get different signs which agrees with my maths. I assume the issue is due to changing from byte to int then to byte again. Next experiment...
public class BitDemo {
public static void main(String[] args) {
byte bb;
byte bc;
int b= 0x80000000;
int c= 0x80000000;
b >>>= 1;
c >>= 1;
bb=(byte)b;
bc=(byte)c;
System.out.println("b >>>= 1 " + bb );
System.out.println("b >>= 1 " + bc );
So in above all maths is int with a final conversion to byte. This gives me zero in both cases.
Signed: Failed final exam and confused.