* > 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 < > 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 > > > > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/algogeeks -~----------~----~----~----~------~----~------~--~---