Re: Private method has been invokated as interface

In program below a private method has been invokated as interface.

It is techically clear, but it is intuitively unclear.

------ foobar.cpp ------
#include <iostream>
using namespace std;

class Base
{
public:
   virtual void foo() = 0;
};

class Derived : public Base
{
private:
   void foo() { cout << "foo(): I am private" << endl; }
};

int main ()
{
   Base* p = new Derived();

   // ----------------------------------------------------------
   // Checks public Base::foo() in compile-time;
   // invokes private Derived::foo() in run-time
   p->foo();
   // ----------------------------------------------------------

   return 0;
}

------ foobar.cpp ------

------ Run ------

$ ./a.exe

foo(): I am private

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