Getting function pointer of private member function in C++ - c++

Question

Is there a way to get function pointer for a member function that is private inside a class
class A
{
public:
void callMe()
{
cout<<__FUNCTION__<<endl;
}
private:
void fooMem()
{
cout<<__FUNCTION__<<endl;
}
};
int _tmain(int argc, _TCHAR* argv[])
{
auto fp = &A::fooMem;
return 0;
}
Compiling this in vs 2012 c++ compiler causes below error
error C2248: 'A::fooMem' : cannot access private member declared in class 'A'
see declaration of 'A::fooMem'
I looked into a amazing solution to a similar problem (though I am not very clear how this actually works, if someone can explain that would be great too), Here I want the address of the member not to invoke it.
The reason I am asking for the address is I'll be patching this function with a different implementation.
The class as such is not modifiable, But I can inherit if that can help achieve this.

Answer 1

Since you cannot modify the original class, it would be much easier to simply inherit the class and create a duplicate of the function in a public memberspace.
I tried this out and it works as expected, at least with g++:
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;
class A
{
public:
void aFunction()
{
cout<<"This is aFunction"<<endl;
}
private:
void anotherFunction()
{
cout<<"This is anotherFunction"<<endl;
}
};
class B: public A
{
public:
void anotherFunction()
{
cout<<"This is also anotherFunction, but it's accessible!"<<endl;
}
};
int main(int argc, char* argv[])
{
A firstClass;
B coach;
firstClass.aFunction();
coach.anotherFunction();
return 0;
}
When I run this code I get the following output:
$ ./a.out
This is aFunction
This is also anotherFunction, but it's accessible!
Proving the compiler understood which version of anotherFunction to use.