shifting problem

[payal sharma]

given
i=-1
-1>>31;
what I have done after reading a post here
first of written one as
00000000 00000000 00000000 00000000 00000001
then I inverted all bits
11111111 11111111 11111111 11111110
then I add one to it
11111111 11111111 11111111 11111111
then when i shifted it to right 31 bit
00000000 00000000 00000000 00000001
now the value is 1
Answer is -1 which is correct
where I made the mistake?

[Desai Sandeep]

Payal,
This is the way you have to go about byte shifting
<pre>
1 ===> 0000 0000 0000 0000 0000 0000 0000 0001
~(inverse) ===> 1111 1111 1111 1111 1111 1111 1111 1110
Add 1 ===> 1
-1 ===> 1111 1111 1111 1111 1111 1111 1111 1111
>> 31 bits ===> 1111 1111 1111 1111 1111 1111 1111 1111
</pre>
Note that, while doing a right shift you have to take fill the places with the sign bit in the left.In this case it is 1 and not 0.
So the result is -1 and not 1.
Hope this helps,
Sandeep
SCJP2, OCSD(Oracle JDeveloper), OCED(Oracle Internet Platform)

[Desai Sandeep]

Rajani,
-1 is not represented as
1000 0000 0000 0000 0000 0000 0000 0001
in binary format!
-- Sandeep

[Mini Pilla]

I am sorry ,I used wrong words!!! Of course that is not a binary represetation but I thought that will help to solve the questions in the exam point of view!!!
Thanx
Rajani

Originally posted by Desai Sandeep:
Rajani,
-1 is not represented as
1000 0000 0000 0000 0000 0000 0000 0001
in binary format!
-- Sandeep

[Ashish Hareet]

Hi Payal ,
Just my addition
When you are shifthing right ( >> ) --- the leftmost(top) bits exposed by the right shift are filled in with the previous contents of the top bit & not just zero's .
We would've got the answer as 1 if we were using " >>> " instead of " >> " .
" >>> " always shifts zero's into the high order bit .
This should help you explain the answer .
The interesting thing about your question is that no matter how much you use " >> " to shift -1 to the right the result would always be -1 .
Hope hat helps
[This message has been edited by Ashish Hareet (edited July 27, 2001).]

[Bill Tripper]

Originally posted by payal sharma:
given
i=-1
-1>>31;

Am I missing something here? The way I figure this is:
i=-1
-1>>31;
is the same as
i=-1 -1>>31;
or
i = (-1) - (1>>31);
or
i = (-1) - 0
or
i = -1
Bill

[Jon Miller]

From the code originally submitted I would have to agree with the last posting!