@pradeep
your ques is not clear..

On Thu, Oct 15, 2009 at 7:58 AM, Arun <[email protected]> wrote:

>
>
> *
>> n=(11101001>>2)|(**11101001<<4)
>> *
>> *n=(00111010)|(01000000)
>>
>> *
>
>
> *11101001<<4 is 10010000*
>
>
>
>> *now n is 01111010*.....
>>
>>
>> On Tue, Oct 13, 2009 at 9:18 AM, Raghavendra Sharma <
>> [email protected]> wrote:
>>
>>> @Ankur I am assuming the integer to be 32 bits. actually it should be
>>> 0xFFFFFFFF
>>> step 1 :  temp =  (0xFFFFFFFF >> (32 - x)) & n;
>>> step 2 :  n  =  (n  >> x) | ( temp << (32 -x));
>>>
>>> The first step extracts the lower x bits and second step moves upper bits
>>> to left side and puts the lower x bits at the beginning.
>>>
>>> for example the integer is  0x12345678 and x = 4 then
>>> temp = 0x8
>>>
>>> (n >> x) = 0x01234567 and temp << (32 - x) is 0x80000000
>>>
>>> and  (n >> x) | (temp << (32 -x)) ix 0x81234567
>>>
>>>
>>> So temp will contain
>>>
>>> On Tue, Oct 13, 2009 at 12:07 AM, GauravNITW <[email protected]>wrote:
>>>
>>>>
>>>> How abt this..?
>>>>
>>>> for(i=0;i<x;i++)
>>>>  {
>>>>    res=no&1U;
>>>>    no=no>>1;
>>>>    if(res==1)
>>>>      no=no|32768U;
>>>>    else
>>>>      no=no|0U;
>>>>  }
>>>>  printf("\nFinal value %u",no);
>>>>
>>>>
>>>> On Oct 12, 8:11 pm, Raghavendra Sharma <[email protected]>
>>>> wrote:
>>>> > temp =  (0xFFFF >> (32 - x)) & n;
>>>> > n  =  (n  >> x) | ( temp << (32 -x));
>>>> >
>>>> > On Mon, Oct 12, 2009 at 5:32 PM, ankur aggarwal <
>>>> [email protected]>wrote:
>>>> >
>>>> >
>>>> >
>>>> > > *You are given a integer and you want to rotate the bits of the
>>>> number by
>>>> > > a value x. Consider the right rotation by x means the least
>>>> significant x
>>>> > > bits should go out from left and take the position of most
>>>> significant x
>>>> > > bits.*- Hide quoted text -
>>>> >
>>>> > - Show quoted text -
>>>>
>>>>
>>>>
>>>
>>>
>>>
>>
>>
>> --
>> Thanks & Regards
>>
>> Umesh kewat
>>
>>
>>
>>
>>
>>
>
> >
>

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