Question

Why is base inaccessible to deriv inside deriv? The program compiles with class deriv : public base.

#include <cstdio>

class base
{
};

class deriv : base
{
  public:
  void f(deriv, int){printf("deriv::f(deriv, int)\n");}
  void f(base){printf("deriv::f(base)\n");}
};

int main()
{
  deriv d;
  d.f(d);
}

17: error: ‘base’ is an inaccessible base of ‘deriv’
17: error:   initializing argument 1 of ‘void deriv::f(base)’

Because two people got it wrong already, I will ask in bold: why does base need to be publicly inherited? It is accessed from within deriv only.

Answer 1

You seem to be incorrectly assuming that conversion from deriv to base when calling deriv::f(base) occurs “inside deriv” and thus has to be accessible. This is not the case. When you call a function, all conversions necessary for initializing function’s arguments occur in the caller’s context. They are not “inside derive“. They happen in the “outside world”. And in your case the “outside world” has no access to deriv-to-base conversion.

In your specific case it is main that is trying to convert deriv to base. And main cannot do it since it has no access to the private base of deriv. Just to experiment you can declare int main() as a friend of deriv and the code will compile.