access private member using template trick

Question

From a blog post Access to private members: Safer nastiness by Johannes Schaub - litb:

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;
}

how get function can be call from a object since its not defined inside class A ?

EDIT:

I don't understand why get must have Tag as parameter instead of a.*get<A_f>() => ok it's due to ADL mechanism

Answer 1

You are not calling get from a! Actually what get return is a class pointer to a member inside A and type of it is int A::* so you need an instance of A to access that value.

For example let me play a little with your code:

struct A {
    A(int a):a(a) { }
    int b;
private:
    int a;
};
void test() {
    auto p = &A::b;
    std::cout << a.*p << std::endl;
}

Did I call p from inside a? a does not have p, this is exactly what happened in your code, get function return &A::a and you use a to read its value! that's all, nothing is wrong and I think it will be compiled in all compilers.

One other question here is: Why C++ allow declaring template using private member of A. C++ standard say:

14.7.2p8 The usual access checking rules do not apply to names used to specify explicit instantiations. [Note: In particular, the template arguments and names used in the function declarator (including parameter types, return types and exception specifications) may be private types or objects which would normally not be accessible and the template may be a member template or member function which would not normally be accessible.]

But if you try to instantiate or even typedef specified template then you get an error. Let's modify your example slightly:

struct A {
private:
    int a;
    friend void f();
};

// Explicit instantiation - OK, no access checks
template struct Rob<A_f, &A::a>;

// Try to use the type in some way - get an error.
struct Rob<A_f, &A::a> r;            // error
typedef struct Rob<A_f, &A::a> R;    // error
void g(struct Rob<A_f, &A::a>);      // error

// However, it's Ok inside a friend function.
void f() {
    Rob<A_f, &A::a> r;               // OK
    typedef Rob<A_f, &A::a> R;       // OK
}

Answer 2

It's legal because friend functions are always in the global scope, even if you implement them inside a class. In other words, this:

class A
{
    friend void go() {}
};

is just a shortcut for:

class A
{
    friend void go();
};

void go() {}

Answer 3

This is a known compiler bug in gcc and was fixed in a later release. See-:

http://gcc.gnu.org/bugzilla/show_bug.cgi?id=41437