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c++ - Allowing access to private members

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.

See Question&Answers more detail:os

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1 Answer

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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.


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