@umesh
gud logic..
On Wed, Oct 14, 2009 at 3:32 PM, umesh kewat <umesh1...@gmail.com> wrote:
> *n=(n>>x)|(n<<(32-x));*
>
> *ex: n=11101001, x=2 suppose here n is 8 bit long*
> *
> n=(11101001>>2)|(11101001<<4)
>
> n=(00111010)|(01000000)
>
> now n is 01111010*.....
>
>
> On Tue, Oct 13, 2009 at 9:18 AM, Raghavendra Sharma <
> raghavendra.vel...@gmail.com> 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 <gauravkis...@gmail.com>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 <raghavendra.vel...@gmail.com>
>>> wrote:
>>> > temp = (0xFFFF >> (32 - x)) & n;
>>> > n = (n >> x) | ( temp << (32 -x));
>>> >
>>> > On Mon, Oct 12, 2009 at 5:32 PM, ankur aggarwal <
>>> ankur.mast....@gmail.com>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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