从嵌套结构调用受保护的基类函数

通常嵌套结构可以访问拥有类public、protected和public成员函数。从嵌套结构中调用基类的 protected 成员函数也没有问题，即以下代码可以正确编译和工作：

#include <iostream>
class Base
{
public:
    Base()
    {}
protected:
    void baseProtectedFunc()
    {
        std::cout << __func__ << "Called for Basen";
    }
};
class Derived : public Base
{
public:
    explicit Derived() : Base()
    {}
    void accessBaseProtectedFuncFromNested()
    {
        Nested myNested( this );
        myNested();
    }
private:
    struct Nested
    {
        explicit Nested( Derived* ptr ) : derived_( ptr ) 
        {}
        void operator()()
        {
            derived_->baseProtectedFunc();
        }
        Derived* derived_;
    };
};
int main( int, char** )
{
    Derived myDerived;
    myDerived.accessBaseProtectedFuncFromNested();
    return 0;
}

现在，请考虑以下代码，该代码使用 mpl::inherit_linearly 生成派生的基类，使用mpl::vector类型：

#include <iostream>
#include <typeinfo>
#include <boost/mpl/vector.hpp>
#include <boost/mpl/inherit.hpp>
#include <boost/mpl/inherit_linearly.hpp>
#include <boost/mpl/for_each.hpp>
template<typename T>
class Base
{
public:
    Base()
    {}
protected:
    void baseProtectedFunc()
    {
        std::cout << __func__ << "Called for Base< " << typeid(T).name() << " >n";
    }
};
typedef boost::mpl::vector< long
    , unsigned
    , bool
    , std::string
    > parameter_type_list_t;
typedef boost::mpl::inherit_linearly< parameter_type_list_t
                  , boost::mpl::inherit< boost::mpl::_1
                                       , Base< boost::mpl::_2 > > 
                  >::type base_types;
class Derived : public base_types
{
public:
    explicit Derived() : base_types()
    {}
    template<typename T>
    void accessBaseProtectedFuncFromNested()
    {
        Nested myNested( this );
        myNested.someFunc<T>();
    }
private:
    struct Nested
    {
        explicit Nested( Derived* ptr ) : derived_( ptr ) 
        {}
        template< typename T >
        void someFunc()
        {
            Base<T>* base = static_cast<Base<T>*>( derived_ );
            base->baseProtectedFunc();
        }
        Derived* derived_;
    };
};
int main( int, char** )
{
    Derived myDerived;
    myDerived.accessBaseProtectedFuncFromNested<unsigned>();
    return 0;
}

使用 GCC 版本 4.4.6-3（在 c++03 和 c++0x 模式下），将生成以下错误：

friend-prot.cpp: In member function ‘void Derived::Nested::someFunc() [with T = unsigned int]’:
friend-prot.cpp:47:   instantiated from ‘void Derived::accessBaseProtectedFuncFromNested() [with T = unsigned int]’
friend-prot.cpp:82:   instantiated from here
friend-prot.cpp:17: error: ‘void Base<T>::baseProtectedFunc() [with T = unsigned int]’ is protected
friend-prot.cpp:72: error: within this context

如果我使我尝试调用的函数public代码将按预期编译和工作。

我可以通过向派生添加一个额外的 private 成员函数来解决这个问题，该函数只是转发来自 Nested 的调用，即：

struct Nested
{
    explicit Nested( Derived* ptr ) : derived_( ptr ) 
    {}
    template< typename T >
    void operator()()
    {
        derived_->forwarder<T>();
    }
    Derived* derived_;
}; 

template< typename T >
void forwarder()
{
    Base<T>::baseProtectedFunc();
}

我不明白为什么我不能调用baseProtectedFunc()，如果在使用mpl::inherit时protected。

为什么允许我在第一个示例中调用基类保护函数，而在第二个示例中则不允许？