C++ inheritance and member function pointers

C++03 std, §4.11 2 Pointer to member conversions:

An rvalue of type “pointer to member of B of type cv T,” where B is a class type, can be converted to an rvalue of type “pointer to member of D of type cv T,” where D is a derived class (clause 10) of B. If B is an inaccessible (clause 11), ambiguous (10.2) or virtual (10.1) base class of D, a program that necessitates this conversion is ill-formed. The result of the conversion refers to the same member as the pointer to member before the conversion took place, but it refers to the base class member as if it were a member of the derived class. The result refers to the member in D’s instance of B. Since the result has type “pointer to member of D of type cv T,” it can be dereferenced with a D object. The result is the same as if the pointer to member of B were dereferenced with the B sub-object of D. The null member pointer value is converted to the null member pointer value of the destination type. ⁵²⁾

⁵²⁾The rule for conversion of pointers to members (from pointer to member of base to pointer to member of derived) appears inverted compared to the rule for pointers to objects (from pointer to derived to pointer to base) (4.10, clause 10). This inversion is necessary to ensure type safety. Note that a pointer to member is not a pointer to object or a pointer to function and the rules for conversions of such pointers do not apply to pointers to members. In particular, a pointer to member cannot be converted to a void*.

In short, you can convert a pointer to a member of an accessible, non-virtual base class to a pointer to a member of a derived class as long as the member isn't ambiguous.

class A {
public: 
    void foo();
};
class B : public A {};
class C {
public:
    void bar();
};
class D {
public:
    void baz();
};
class E : public A, public B, private C, public virtual D {
public: 
    typedef void (E::*member)();
};
class F:public E {
public:
    void bam();
};
...
int main() {
   E::member mbr;
   mbr = &A::foo; // invalid: ambiguous; E's A or B's A?
   mbr = &C::bar; // invalid: C is private 
   mbr = &D::baz; // invalid: D is virtual
   mbr = &F::bam; // invalid: conversion isn't defined by the standard
   ...

Conversion in the other direction (via static_cast) is governed by § 5.2.9 9:

An rvalue of type "pointer to member of D of type cv1 T" can be converted to an rvalue of type "pointer to member of B of type cv2 T", where B is a base class (clause 10 class.derived) of D, if a valid standard conversion from "pointer to member of B of type T" to "pointer to member of D of type T" exists (4.11 conv.mem), and cv2 is the same cv-qualification as, or greater cv-qualification than, cv1.¹¹⁾ The null member pointer value (4.11 conv.mem) is converted to the null member pointer value of the destination type. If class B contains the original member, or is a base or derived class of the class containing the original member, the resulting pointer to member points to the original member. Otherwise, the result of the cast is undefined. [Note: although class B need not contain the original member, the dynamic type of the object on which the pointer to member is dereferenced must contain the original member; see 5.5 expr.mptr.oper.]

¹¹⁾ Function types (including those used in pointer to member function types) are never cv-qualified; see 8.3.5 dcl.fct.

In short, you can convert from a derived D::* to a base B::* if you can convert from a B::* to a D::*, though you can only use the B::* on objects that are of type D or are descended from D.

I'm not 100% sure what you are asking, but here is an example that works with virtual functions:

#include <iostream>
using namespace std;

class A { 
public:
    virtual void foo() { cout << "A::foo\n"; }
};
class B : public A {
public:
    virtual void foo() { cout << "B::foo\n"; }
};

int main()
{
    void (A::*bar)() = &A::foo;
    (A().*bar)();
    (B().*bar)();
    return 0;
}

The critical issue with pointers to members is that they can be applied to any reference or pointer to a class of the correct type. This means that because Z is derived from Y a pointer (or reference) of type pointer (or reference) to Y may actually point (or refer) to the base class sub-object of Z or any other class derived from Y.

void (Y::*p)() = &Z::z_fn; // illegal

This means that anything assigned to a pointer to member of Y must actually work with any Y. If it was allowed to point to a member of Z (that wasn't a member of Y) then it would be possible to call a member function of Z on some thing that wasn't actually a Z.

On the other hand, any pointer to member of Y also points the member of Z (inheritance means that Z has all the attributes and methods of its base) is it is legal to convert a pointer to member of Y to a pointer to member of Z. This is inherently safe.

void (Y::*p)() = &Y::y_fn;
void (Z::*q)() = p; // legal and safe

You might want to check out this article Member Function Pointers and the Fastest Possible C++ Delegates The short answer seems to be yes, in some cases.

I believe so. Since the function pointer uses the signature to identify itself, the base/derived behavior would rely on whatever object you called it on.