*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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