@saurabh u are getting sizeof(a)/sizeofa[0] =1 coz fiest one is pointer and second one is integer...both's size is 4 do it without passing http://www.ideone.com/8olTP
On Tue, Aug 23, 2011 at 1:28 PM, vikas <vikas.rastogi2...@gmail.com> wrote: > nopes, you need to know where the hell it ends.... even if this is a > string , it ends with convention of ending 0. in case it is stream , > we know the data length. in case of array, above mentioned approach > should work. sizeof(arr)/sizeof(arr[0]) > > if you are given only a pointer and no length, you can address until > there is another page starts in memory , not belonging to the process. > > > > > On Aug 23, 7:07 am, saurabh singh <saurab...@gmail.com> wrote: > > Just a small code to back up my point...http://www.ideone.com/woRiT > > > > > > > > > > > > > > > > > > > > On Tue, Aug 23, 2011 at 7:33 AM, saurabh singh <saurab...@gmail.com> > wrote: > > > That would take all the fun away....what if you are given only the > address > > > of the array?This wont work in that case > > > > > On Mon, Aug 22, 2011 at 10:39 PM, asdqwe <ayushgoel...@gmail.com> > wrote: > > > > >> If i am not wrong, the only possible solution can be > > >> len=sizeof(arr)/sizeof(arr[0]) > > >> i.e. find the length from the array itself. > > > > >> On Aug 22, 9:01 pm, saurabh singh <saurab...@gmail.com> wrote: > > >> > @dave or anyone??????? response please > > > > >> > On Sun, Aug 21, 2011 at 12:43 PM, saurabh singh < > saurab...@gmail.com> > > >> wrote: > > >> > > kkk...not sure > > >> > > assume no number is greater than 1000(I mentioned There has to be > some > > >> > > additional constraints to make the problem solvable).... > > >> > > Now check 1st element if not the desired element keep multiplying > with > > >> 2 > > >> > > the previous range till either one of these condition is satisfied > > >> > > *1.An exception is caught* > > >> > > *2.Number greater than 1000 occurs.* > > >> > > suppose this happens for *1024 *for the given example. > > >> > > then we will check out for (512+1024)/2 th element for the above > > >> condition. > > >> > > If true than again branch like binary search.This way can element > > >> which on > > >> > > left side doesn't gives any exception and maintains the > constraints > > >> while on > > >> > > the right it violates the same.So we may land up with the desired > > >> index and > > >> > > can then perform binary search....... > > > > >> > > PS:There are lots of assumption in this approach and the more I > write > > >> the > > >> > > more I get convinced that its a plain stupid idea... > > > > >> > > -- > > >> > > Saurabh Singh > > >> > > B.Tech (Computer Science) > > >> > > MNNIT ALLAHABAD > > > > >> > -- > > >> > Saurabh Singh > > >> > B.Tech (Computer Science) > > >> > MNNIT ALLAHABAD > > > > >> -- > > >> 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?hl=en. > > > > > -- > > > Saurabh Singh > > > B.Tech (Computer Science) > > > MNNIT ALLAHABAD > > > > -- > > Saurabh Singh > > B.Tech (Computer Science) > > MNNIT ALLAHABAD > > -- > 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?hl=en. > > -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- 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?hl=en.
