Very True. Experience counts on each field. ;)

K.Sant.

hi , i am trying to get the object name, see the code below, but not able to find method, what could be the code in place of ???????. #include "iostream" using namespace std; class A{ public: int i; void printObjectname(){ cout << ????????;//<--what could be here, to print the object name as 'objectA' } }; int _tmain(int argc, _TCHAR* argv[]) { A *objectA = new A(); objectA->printObjectname(); return 0; } santosh

saayed 1

saayed