Question

Please check this code..

class ex
{
     int i;
     public:
     ex(int ii = 0):i(ii){}
      ~ex(){cout<<"dest"<<endl;}
     void show()
     {
        cout<<"show fun called"<<endl;
     }
};
int main(int argc , char *argv[])
{
    ex *ob = NULL;
     ob->show();
       return 0;
}

what happens when we call show method.

Thanks..

Answer 1

ex *ob = NULL;
ob->show();

You’re dereferencing a null pointer which causes undefined behaviour. This is bad.

If it’s not clear where the dereference is then understand that the -> operator translates to
(*ob).show().