depends upon the optimization performed. Some compilers, on seeing a const value, store there literal value to get faster access (as it is assumed that const won't be modified) Hence you get the output:0 2 as value of i is now the already stored literal in the optimized plan generated by the compiler.
But if such an optimization is not performed, then you get 2,2 On 22 August 2011 00:03, Abhishek <[email protected]> wrote: > i am compiling it on linux, it is giving me the o/p 2,2. > > whatever i know is that, to prevent constants being modified, pointers > should also be declared constant. > like.. > const int *p; > > now it will produce error. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To view this discussion on the web visit > https://groups.google.com/d/msg/algogeeks/-/QBiQoXPIk78J. > > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- Regards Siddharth Srivastava -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
