Nested Class member function can't access function of enclosing class. Why?

Question

Please see the example code below:

class A
{
private:
    class B
    {
    public:
        foobar();
    };
public:
    foo();
    bar();
};

Within class A & B implementation:

A::foo()
{
    //do something
}

A::bar()
{
    //some code
    foo();
    //more code
}

A::B::foobar()
{
    //some code
    foo(); //<<compiler doesn't like this
}

The compiler flags the call to foo() within the method foobar(). Earlier, I had foo() as private member function of class A but changed to public assuming that B's function can't see it. Of course, it didn't help. I am trying to re-use the functionality provided by A's method. Why doesn't the compiler allow this function call? As I see it, they are part of same enclosing class (A). I thought the accessibility issue for nested class meebers for enclosing class in C++ standards was resolved.

How can I achieve what I am trying to do without re-writing the same method (foo()) for B, which keeping B nested within A?

I am using VC++ compiler ver-9 (Visual Studio 2008). Thank you for your help.

Answer 1

foo() is a non-static member function of A and you are trying to call it without an instance.
The nested class B is a seperate class that only has some access privileges and doesn't have any special knowledge about existing instances of A.

If B needs access to an A you have to give it a reference to it, e.g.:

class A {
    class B {
        A& parent_;
    public:
        B(A& parent) : parent_(parent) {}
        void foobar() { parent_.foo(); }
    };
    B b_;
public:
    A() : b_(*this) {}
};

Answer 2

This is an automagic, albeit possibly nonportable trick (worked on VC++ since 6.0 though). Class B has to be a member of class A for this to work.

#ifndef OUTERCLASS
#define OUTERCLASS(className, memberName) \
    reinterpret_cast<className*>(reinterpret_cast<unsigned char*>(this) - offsetof(className, memberName))
#endif 

class A
{
private:
    class B
    {
    public:
        void foobar() {
           A* pA = OUTERCLASS(A, m_classB);
           pA->foo();
        }
    } m_classB;
public:
    foo();
    bar();
};

Answer 3

If you want to reuse functionality from A then you should inherit from A not nest B inside it.

Answer 4

Basically what Georg Fritzsche said

#include <iostream>
#include <cstring>
using namespace std;

class A
{
private:
    class B
    {
     A& parent_;
     public:
        //B();  //uncommenting gives error
        ~B();
        B(A& parent) : parent_(parent) {}

        void foobar() 
        { 
         parent_.foo();  
         cout << "A::B::foo()" <<endl; 
        }

        const std::string& foobarstring(const std::string& test) const 
        { 
         parent_.foostring(test); cout << "A::B::foostring()" <<endl;
        }
    };
public:
    void foo();
    void bar();
    const std::string& foostring(const std::string& test) const;
    A(); 
    ~A(){};
    B b_;
};

//A::B::B() {}; //uncommenting gives error
A::B::~B(){};

A::A():b_(*this) {}


void A::foo()
{
    cout << "A::foo()" <<endl;
}

const std::string& A::foostring(const std::string& test) const
{
    cout << test <<endl;
    return test;
}

void A::bar()
{
    //some code
    cout << "A::bar()" <<endl;
    foo();
    //more code
}

int main(int argc, char* argv[])
{
A a;
a.b_.foobar();
a.b_.foobarstring("hello");

return 0;
}

If you uncomment the default B constructor you would get an error