Allowing access to private members

问题:

This question is somewhat a continuation of this one I've posted.

What I was trying to do: my point was to allow access to private members of a base class A in a derived class B, with the following restraints:

• what I want to access is a structure -- an std::map<>, actually --, not a method;
• I cannot modified the base class;
• base class A has no templated method I may overload as a backdoor alternative -- and I would not add such method, as it would be going against the second restraint.

As a possible solution, I have been pointed to litb's solution (post / blog), but, for the life of me, I haven't been able to reach an understanding on what is done in these posts, and, therefore, I could not derive a solution to my problem.

What I am trying to do: The following code, from litb's solution, presents an approach on how to access private members from a class/struct, and it happens to cover the restrictions I've mentioned.

So, I'm trying to rearrange this one code:

template<typename Tag, typename Tag::type M>
struct Rob { 
  friend typename Tag::type get(Tag) {
    return M;
  }
};

// use
struct A {
  A(int a):a(a) { }
private:
  int a;
};

// tag used to access A::a
struct A_f { 
  typedef int A::*type;
  friend type get(A_f);
};

template struct Rob<A_f, &A::a>;

int main() {
  A a(42);
  std::cout << "proof: " << a.*get(A_f()) << std::endl;
}

To something that would allow me to do the following -- note I'm about to inherit the class, as the entries in the std::map<> are added right after the initialization of the derived class B, i.e., the std::map<> isn't simply a static member of class A with a default value, so I need to access it from a particular instance of B:

// NOT MY CODE -- library <a.h>
class A {
private:
    std::map<int, int> A_map;
};

// MY CODE -- module "b.h"
# include <a.h>
class B : private A {
public:
    inline void uncover() {
        for (auto it(A_map.begin()); it != A_map.end(); ++it) {
            std::cout << it->first << " - " << it->second << std::endl;
        }
    }
};

What I'd like as an answer: I'd really love to have the above code working -- after the appropriate modifications --, but I'd be very content with an explanation on what is done in the first code block -- the one from litb's solution.

回答1:

The blog post and its code is unfortunately a bit unclear. The concept is simple: an explicit template instantiation gets a free backstage pass to any class, because

• An explicit instantiation of a library class may be an implementation detail of a client class, and
• Explicit instantiations may only be declared at namespace scope.

The natural way to distribute this backstage pass is as a pointer to member. If you have a pointer to a given class member, you can access it in any object of that class regardless of the access qualification. Fortunately, pointer-to-members can be compile-time constants even in C++03.

So, we want a class which generates a pointer to member when it's explicitly instantiated.

Explicit instantiation is just a way of defining a class. How can merely generating a class do something? There are two alternatives:

• Define a friend function, which is not a member of the class. This is what litb does.
• Define a static data member, which gets initialized at startup. This is my style.

I'll present my style first, then discuss its shortcoming, and then modify it to match litb's mechanism. The end result is still simpler than the code from the blog.

Simple version.

The class takes three template arguments: the type of the restricted member, its actual name, and a reference to a global variable to receive a pointer to it. The class schedules a static object to be initialized, whose constructor initializes the global.

template< typename type, type value, type & receiver >
class access_bypass {
    static struct mover {
        mover()
            { receiver = value; }
    } m;
};

template< typename type, type value, type & receiver >
typename access_bypass< type, value, receiver >::mover
    access_bypass< type, value, receiver >::m;

Usage:

type_of_private_member target::* backstage_pass;
template class access_bypass <
    type_of_private_member target::*,
    & target::member_name,
    backstage_pass
>;

target t;
t.* backstage_pass = blah;

See it work.

Unfortunately, you can't rely on results from this being available for global-object constructors in other source files before the program enters main, because there's no standard way to tell the compiler which order to initialize files in. But globals are initialized in the order they're declared, so you can just put your bypasses at the top and you'll be fine as long as static object constructors don't make function calls into other files.

Robust version.

This borrows an element from litb's code by adding a tag structure and a friend function, but it's a minor modification and I think it remains pretty clear, not terribly worse than the above.

template< typename type, type value, typename tag >
class access_bypass {
    friend type get( tag )
        { return value; }
};

Usage:

struct backstage_pass {}; // now this is a dummy structure, not an object!
type_of_private_member target::* get( backstage_pass ); // declare fn to call

// Explicitly instantiating the class generates the fn declared above.
template class access_bypass <
    type_of_private_member target::*,
    & target::member_name,
    backstage_pass
>;

target t;
t.* get( backstage_pass() ) = blah;

See it work.

The main difference between this robust version and litb's blog post is that I've collected all the parameters into one place and made the tag structure empty. It's just a cleaner interface to the same mechanism. But you do have to declare the get function, which the blog code does automatically.

回答2:

OK, so you asked about how to make that weird "Rob" code work with your use case, so here it is.

// the magic robber
template<typename Tag, typename Tag::type M>
struct Rob {
    friend typename Tag::type get(Tag) {
        return M;
    }
};

// the class you can't modify
class A {
private:
    std::map<int, int> A_map;
};

struct A_f {
    typedef std::map<int, int> A::*type;
    friend type get(A_f);
};

template struct Rob<A_f, &A::A_map>;

class B : private A {
public:
    inline void uncover() {
        std::map<int, int>::iterator it = (this->*get(A_f())).begin();
    }
};

Now, I personally think the cure here may be worse than the disease, despite that I'm usually the last one you'll see claiming that abusing C++ is OK. You can decide for yourself, so I've posted this as a separate answer from my one using the preprocessor to do it the old-school way.

Edit:

How It Works

Here I will replicate the code above, but with the types simplified and the code drawn out more, with copious comments. Mind you, I did not understand the code very well before I went through this exercise, I don't understand it completely now, and I certainly won't remember how it works tomorrow. Caveat maintainer.

Here's the code we aren't allowed to change, with the private member:

// we can use any type of value, but int is simple
typedef int value_type;

// this structure holds value securely.  we think.
struct FortKnox {
    FortKnox() : value(0) {}
private:
    value_type value;
};

Now for the heist:

// define a type which is a pointer to the member we want to steal
typedef value_type FortKnox::* stolen_mem_ptr;

// this guy is sort of dumb, but he knows a backdoor in the standard
template<typename AccompliceTag, stolen_mem_ptr MemPtr>
struct Robber {
    friend stolen_mem_ptr steal(AccompliceTag) {
        return MemPtr; // the only reason we exist: to expose the goods
    }
};

// this guy doesn't know how to get the value, but he has a friend who does
struct Accomplice {
    friend stolen_mem_ptr steal(Accomplice);
};

// explicit instantiation ignores private access specifier on value
// we cannot create an object of this type, because the value is inaccessible
// but we can, thanks to the C++ standard, use this value in this specific way
template struct Robber<Accomplice, &FortKnox::value>;

// here we create something based on secure principles, but which is not secure
class FortKnoxFacade : private FortKnox {
public:
    value_type get_value() const {
        // prepare to steal the value
        // this theft can only be perpetrated by using an accomplice
        stolen_mem_ptr accessor = steal(Accomplice()); // it's over now
        // dereference the pointer-to-member, using this as the target
        return this->*accessor;
    }
};

int main() {
    FortKnoxFacade fort;
    return fort.get_value();
}

回答3:

How about something more brutal?

// MY CODE -- module "b.h"
# define private protected
# include <a.h>
# undef private
class B : private A {
    // now you can access "private" members and methods in A

回答4:

The best-packaged version I know of this idiom is as follows:

template<class Tag,typename Tag::type MemberPtr>
struct access_cast{
 friend typename Tag::type get(Tag){return MemberPtr;};
};

template<class Tag,class MemberPtr>
struct access_tag{
 typedef MemberPtr type;
 friend type get(Tag);
};

class A {
public:
 auto x() const {return x_;};
private: 
 int x_ = 9;
};

#include <iostream>

struct AMemTag: access_tag<AMemTag,int A::*>{}; //declare tag
template struct access_cast<AMemTag,&A::x_>; //define friend get function

int main() {
 A a;
 std::cout<<a.x()<<"\n";
 a.*get(AMemTag()) = 4; //dereference returned member pointer and modify value
 std::cout<<a.x()<<"\n";
}

See it work.