shift operators

[nitin sharma]

1) public class x
2) {
3) public static void main(String[]args)
4) {
5) try
6) {
7) int i=(-1>>65535)+1;
8) int j=5/i;
9) System.out.println("end of try block");
10) }
11) catch(Exception e)
12) {
13) System.exit(0);
14) }
15) finally
16) {
17) System.out.println("finally");
18) }
}
}
Can anybody please workout on the line number 7.Hope i will get a sooner reply.

[Ajith Kallambella]

When the number of positions being shifted exceeds the length of the datatype( in bytes ), the actual shift occurs only by ( numPositions % datatypeSizeInBytes ) positions.
ie.,
when you shift an integer by X positions where X is greater than 32, the actual positions shifted is not X, but ( X % 32 )
Similarly, when you shift a long by X positions where X is greater than 64, the actual positions shifted is not X, but ( X % 64 ).
Now lets look at the line
int i=(-1>>65535)+1;
Since 65535 is greater than 32, you will first have to figure out the actual number of positions for the shift. Use the formula mentioned above.
int i=(-1>>(65535%32)) + 1;
ie., int i=( -1>>31 ) + 1;
-1 >> 31 is -1 so the answer is 0.
Cheers!
------------------
Ajith Kallambella M.
Sun Certified Programmer for the Java�2 Platform.
IBM Certified Developer - XML and Related Technologies, V1.
[This message has been edited by Ajith Kallambella (edited May 02, 2001).]

[Lam Thai]

Originally posted by nitin sharma:
1) public class x
2) {
3) public static void main(String[]args)
4) {
5) try
6) {
7) int i=(-1>>65535)+1;
8) int j=5/i;
9) System.out.println("end of try block");
10) }
11) catch(Exception e)
12) {
13) System.exit(0);
14) }
15) finally
16) {
17) System.out.println("finally");
18) }
}
}
Can anybody please workout on the line number 7.Hope i will get a sooner reply.

Hi Nitin,
Signed shift right of -1 always yield -1 regardless how many times it shifts.
- Lam

[shadow liu]

hi, Ajith,
Everything you said is right, except 65535%32 should be 31

[nitin sharma]

Hi Ajith,
What role does the +1 play's in finding the bit's shifted in line number 7.Please explain it.

[Ravindra Mohan]

Hi Nitin,
Well the answer to your querry is simple it just adds 1 to the result of shifting the signed right shift >> bits of '-1'.
Please note that -1 is the only number which when <code> >> </code> shifted will always give the same result.
There is no mystery in this, the reason is very simple if you consider the fantastic bit
value of -1 which is = <code> 1111 1111 1111 1111 1111 1111 1111 1111 </code>.

Now let us right shift -1 by 2 bits so we have ??11 1111 1111 1111 1111 1111 1111 1111 . But in case of signed right shift <code> >> </code>. The ?? are replaced by two 1's (also known as parity bits). Thus result is 1111 1111 1111 1111 1111 1111 1111 1111 which is again -1.
Thus if you see that in case of -1 whatever times we right shift we get the same result back.
Cheers,
Ravindra Mohan
[This message has been edited by Ravindra Mohan (edited May 03, 2001).]