Calling private method in C++

This is purely a theoretical question, I know that if someone declares a method private, you probably shouldn't call it. I managed to call private virtual methods and change private members for instances, but I can't figure out how to call a private non-virtual method (without using __asm). Is there a way to get the pointer to the method? Are there any other ways to do it?

EDIT: I don't want to change the class definition! I just want a hack/workaround. :)

Solution 1:

See my blog post. I'm reposting the code here

template<typename Tag>
struct result {
  /* export it ... */
  typedef typename Tag::type type;
  static type ptr;
};

template<typename Tag>
typename result<Tag>::type result<Tag>::ptr;

template<typename Tag, typename Tag::type p>
struct rob : result<Tag> {
  /* fill it ... */
  struct filler {
    filler() { result<Tag>::ptr = p; }
  };
  static filler filler_obj;
};

template<typename Tag, typename Tag::type p>
typename rob<Tag, p>::filler rob<Tag, p>::filler_obj;

Some class with private members

struct A {
private:
  void f() {
    std::cout << "proof!" << std::endl;
  }
};

And how to access them

struct Af { typedef void(A::*type)(); };
template class rob<Af, &A::f>;

int main() {
  A a;
  (a.*result<Af>::ptr)();
}

Solution 2:

#include the header file, but:

#define private public
#define class struct

Clearly you'll need to get around various inclusion guards etc and do this in an isolated compilation unit.

EDIT: Still hackish, but less so:

#include <iostream>

#define private friend class Hack; private

class Foo
{
public:
    Foo(int v) : test_(v) {}
private:
    void bar();
    int test_;
};
#undef private
void Foo::bar() { std::cout << "hello: " << test_ << std::endl; }

class Hack
{
public:
    static void bar(Foo& f) {
        f.bar();
    }
};

int _tmain(int argc, _TCHAR* argv[])
{
    Foo f(42);
    Hack::bar(f);
    system("pause");
    return 0;
}