@Lego: I am sorry I missed the "address of" operator... I wanted to type: printf("%d", *(char *) &(0x00000002))
But even the above is incorrect since it is not possible to take address of a literal in C/C++. The best way is: const int i=0x00000002; printf("%d", *(char *) &(i)); If this print 2, then its little-endian else big-endian (there exists mixed-endianness also, but lets leave that for now.) --Sundeep. On Tue, Jun 15, 2010 at 9:05 PM, Lego Haryanto <legoharya...@gmail.com>wrote: > > > On Mon, Jun 14, 2010 at 5:13 AM, Sundeep Singh <singh.sund...@gmail.com>wrote: > >> @saurav: your code will always print 2 irrespective of the system's >> endianness! >> >> correct thing to do is: >> printf("%d", *(char *) (0x00000002)) >> >> --Sundeep. >> >> >> > ... dereferencing a very low address pointer, are you sure? > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algoge...@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com<algogeeks%2bunsubscr...@googlegroups.com> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algoge...@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.