C++::Access to more than one private members using templates trick - c++

Question

I am studying the following post access private member using template trick.
I am wondering how the code should be modified in order to access more than one private varible
I tried the following
#pragma once
template<typename Tag, typename Tag::type M>
struct Rob {
friend typename Tag::type get( typename Tag::type) {
return M;
}
};
// use
struct A {
A(int a) :a(a) { }
private:
int a;
int b;
};
// tag used to access A::a
template<typename Tag, typename Member>
struct TagBase {
typedef Member type;
friend type get(Tag);
};
struct A_f : TagBase<A_f, int A::*> { };
template struct Rob<A_f, &A::a>;
template struct Rob<A_f, &A::b>;
int main() {
A a(42);
std::cout << "proof: " << a.*get(A_f()) << std::endl;
}
But I get the following error
error C2084: function 'int A::* Rob<A_f,pointer-to-member(0x0)>::get(int A::* )' already has a body
message : see previous definition of 'get'
message : see reference to class template instantiation 'Rob<A_f,pointer-to-member(0x4)>' being compiled
This the link to the demo

Answer 1

This is because typename Tag::type is both int A::*, so both instantiation define the same function.
To fix this, you'll need to change the example a bit so it uses multiple tag types:
#include <iostream>
template<typename Tag, typename Tag::type M>
struct Rob {
// Here we receive tag directly
friend typename Tag::type get(Tag) {
return M;
}
};
// use
struct A {
A(int a) :a(a) { }
private:
int a;
int b;
};
// tag used to access A::a
template<typename Tag, typename Member>
struct TagBase {
typedef Member type;
friend type get(Tag);
};
struct A_af : TagBase<A_af, int A::*> { };
struct A_bf : TagBase<A_bf, int A::*> { };
template struct Rob<A_af, &A::a>;
template struct Rob<A_bf, &A::b>;
int main() {
A a(42);
std::cout << "proof: " << a.*get(A_bf()) << a.*get(A_af()) << std::endl;
}