Why is it impossible to get a pointer to a protected method of the base class? [duplicate]

Question

class A {
public:
   A() {auto tmp = &A::foo;}
protected:
   void foo() {}
};

class B : public A {
public:
   B() {auto tmp = &A::foo;}
};

Class A compiles no problem. Class B yields a compilation error:

'A::foo' : cannot access protected member declared in class 'A'

Why is that, what's the rationale? Is there a way to circumvent this (if I need that pointer for a callback, std::function etc.)?

Answer 1

Why is that, what's the rationale?

In general, a derived class can only access protected members of the same derived class, not of the base class itself or arbitrary classes with the same base class. So B can access protected members of A via B, but not directly via A.

Is there a way to circumvent this?

The inherited member is also a member of B, and can be accessed as such:

auto tmp = &B::foo;