How to hide a datum from everyone but class T - c++

Question

I want a type A that will yield its hidden datum to an object of type T but hide the datum from everyone else. My C++ compiler happens to be GCC 4.4, but that shouldn't matter. Why won't this work?
#include <iostream>
template <class T> class A {
private:
int n1;
public:
friend class T;
A(const int n0 = 0) : n1(n0) {}
};
class B {
public:
int f(const A<B> a) const { return a.n1; }
B() {}
};
int main() {
const A<B> a(5);
const B b;
const int m = b.f(a);
std::cout << m << "\n";
return 0;
}
Incidentally, this works fine, except that it fails to hide the datum:
#include <iostream>
template <class T> class A {
private:
int n1;
public:
int n() const { return n1; }
A(const int n0 = 0) : n1(n0) {}
};
class B {
public:
int f(const A<B> a) const { return a.n(); }
B() {}
};
int main() {
const A<B> a(5);
const B b;
const int m = b.f(a);
std::cout << m << "\n";
return 0;
}
Does C++ really not allow a friend class to be specified at compile time as a template parameter? Why not? If not, then what alternate technique should I use to hide the datum? (One would prefer a compile-time technique if possible.)
What is my misunderstanding here, please?
(I see some answers to related questions here and here, but either they don't answer my particular question or I fail to understand that they do so. At any rate, maybe I am using the wrong technique altogether. Though I remain interested in why the friend class T fails, what I really want to know is how to hide the datum, whether with a friend or by other means.)
Thanks.

Answer 1

Your compiler is simply too old. C++11 allows you to declare template parameters as friends.
§11.3 [class.friend] p3
A friend declaration that does not declare a function shall have one of the following forms:
friend elaborated-type-specifier ;
friend simple-type-specifier ;
friend typename-specifier ;
If the type specifier in a friend declaration designates a (possibly cv-qualified) class type, that class is declared as a friend; otherwise, the friend declaration is ignored.
And it even contains an example of a template parameter as a friend:
class C;
// [...]
template <typename T> class R {
friend T;
};
R<C> rc; // class C is a friend of R<C>
R<int> ri; // OK: "friend int;" is ignored
C++03 sadly has no way to do this, however you can simply friend a single free function and let that act as "glue" code that takes the data from one class and passes it to the other. Another way might be the passkey pattern.

Answer 2

I don't know the standardese behind your error (refer to Xeo's answer), but I did find a workaround for C++03.
Instead of making T a friend, make one of T's member functions a friend:
#include <iostream>
template <class T> class A {
private:
int n1;
public:
friend int T::getN1(const A& a) const;
A(const int n0 = 0) : n1(n0) {}
};
class B {
public:
int f(const A<B> a) const { return getN1(a); }
B() {}
private:
int getN1(const A<B>& a) const {return a.n1;}
};
class C {
public:
int f(const A<B> a) const { return getN1(a); }
C() {}
private:
// Error, n1 is a private member of A<B>
int getN1(const A<B>& a) const {return a.n1;}
};
int main() {
const A<B> a(5);
const B b;
const int m = b.f(a);
std::cout << m << "\n";
return 0;
}
Alternatively, you can make a nested class/struct of T be a friend of A. This may be more convenient if there are several private members of A that you want T to have access to.
#include <iostream>
template <class T> class A {
private:
int n1;
public:
friend class T::AccessToA;
A(const int n0 = 0) : n1(n0) {}
};
class B {
public:
int f(const A<B> a) const { return AccessToA::getN1(a); }
B() {};
private:
friend class A<B>;
struct AccessToA
{
static int getN1(const A<B>& a) {return a.n1;}
};
};
class C {
public:
int f(const A<B> a) const { return AccessToA::getN1(a); }
C() {};
private:
friend class A<C>;
struct AccessToA
{
// Error, n1 is a private member of A<B>
static int getN1(const A<B>& a) {return a.n1;}
};
};
int main() {
const A<B> a(5);
const B b;
const int m = b.f(a);
std::cout << m << "\n";
return 0;
}