why can't I cast a pointer to Derived class member function to the same but of class Base?

Question

To me it looks perfectly safe to cast a void(Derived::*)() to a void(Base::*)(), like in this code:

#include <iostream>
#include <typeinfo>
using namespace std;
struct Base{
    void(Base::*any_method)();
    void call_it(){
        (this->*any_method)();
    }
};
struct Derived: public Base{
    void a_method(){
        cout<<"method!"<<endl;
    }
};
int main(){
    Base& a=*new Derived;
    a.any_method=&Derived::a_method;
    a.call_it();
}

But the compiler complains about the cast at a.any_method=&Derived::a_method;. Is this a roadblock to prevent subtle programming errors, or just something to make life easier for compiler writers? Are there workarounds to let the Base class have a pointer to member functions of Derived without type knoweledge (that is, I cannot make Base a template with template argument Derived).

Answer 1

What happens if your Derived::a_method() attempts to use a data member only present in Derived, not in Base, and you call it on a Base object (or an object derived from Base but not related to Derived)?

The conversion the other way around makes sense, this one doesn't.