How can I access a protected constructor from a friend function?

Question

I created a class and I want to force anyone who's trying to construct an object, to use unique_ptr . To do that I thought of declaring the constructor protected and use a friend function that returns a unique_ptr . So here's an example of what I want to do:

template <typename T>
class A
{
public:
    friend std::unique_ptr<A<T>> CreateA<T>(int myarg);

protected:
    A(int myarg) {}
};

template <typename T>
std::unique_ptr<A<T>> CreateA(int myarg)
{
    // Since I declared CreateA as a friend I thought I
    // would be able to do that
    return std::make_unique<A<T>>(myarg);
}

I did some reading on friend functions and I understood that a friend function provides access to private/protected members of an object of a class.

---

Is there anyway I can make my example work?

Even without friend functions, my goal is to make the CreateA the only way for someone to create an object.

EDIT

I change the code a bit. I didn't mention that my class takes one template parameter. That makes things more complex apparently.

Answer 1

You can do it this way :-

#include <iostream>
#include <memory>
using namespace std;
class A
{
    int arg;
public:
    friend unique_ptr<A> CreateA(int myarg);
    void showarg() { cout<<arg; }

protected:
    A(int myarg): arg(myarg) {}
};
unique_ptr<A> CreateA (int myarg)
{
    return std::unique_ptr<A>(new A(myarg));
}
int main()
{
    int x=5;
    unique_ptr<A> u = CreateA(x);
    u->showarg();
    return 0;
}

Output :-

5

If you don't want to use friend function you can make the function static & call it like this :-

unique_ptr<A> u = A::CreateA(x);

EDIT :-

In reply to your edit I rewrote the program & it goes like this :-

#include <iostream>
#include <memory>
using namespace std;
template <typename T>
class A
{
    T arg;
public:
    static std::unique_ptr<A> CreateA(T myarg)
    {
        return std::unique_ptr<A>( new A(myarg) );
    }
    void showarg()
    {
        cout<<arg;
    }
protected:
    A(T myarg): arg(myarg) {}
};

int main()
{
    int x=5;
    auto u = A<int>::CreateA(x);
    u->showarg();
    return 0;
}

Simple & easy !!! But remember you cannot instantiate the object of A . Good Luck !!!

Answer 2

Create a static function that instantiates the protected constructor.

  #include<iostream>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include <memory>
using namespace std;
template< typename T >
class A
{
public:
    static void CreateA(int myarg, std::unique_ptr<A<T>>& objA, T t) {
        std::unique_ptr<A<T>> objB(new A(myarg, t));
        objA = std::move(objB);
    }

protected:
    A(int myarg, T t) {
        m_t = t;
    }

private:
    T m_t;
};

int main() {

    int myArg = 0;
    std::unique_ptr<A<int>> anotherObjA;
    A<int>::CreateA(myArg, anotherObjA, myArg);


    return 0;
}

Answer 3

The other answers suggest using a static template function, which I agree is the best solution, because it is simpler.

My answer explains why your friend approach didn't work and how to use the friend approach correctly.

---

There are two problems in your original code. One is that make_unique is not actually a friend of A , so the call make_unique<A<T>>(myarg); does not have access to A 's protected constructor. To avoid this , you can use unique_ptr<A<T>>(new A(myarg)) instead. Theoretically it would be possible to declare make_unique a friend but I'm not even sure of the right syntax for that.

The other issue is the template friends problem . Inside a class template, friend <function-declaration> actually declares a non-template friend.

The C++ FAQ suggests two possible workarounds. One of them is to define the friend function inline. However, in that case the function can only be found by argument-dependent lookup. But since the function does not take A<T> (or A<T> & ) as argument, it can never be found this way. So this option is not viable to your situation -- it's more suited to operator overloading.

So the only fix is to declare (and optionally define) the template function before the class definition:

#include <memory>

template<typename T>
class A;

template <typename T>
std::unique_ptr<A<T>> CreateA(int myarg)
{
    return std::unique_ptr<A<T>>{new A<T>(myarg)};
}

template <typename T>
class A
{
    friend std::unique_ptr<A<T>> CreateA <> (int myarg);
//          refers to existing template  ^^

protected:
    A(int myarg) {}
};

int main()
{
    auto x = CreateA<int>(5);
}

Note: It is possible to declare CreateA where I have defined it, and put the function definition later. However, the code I have posted works -- despite A not being defined when new A<T>(myarg) appears in the source -- because CreateA is not instantiated until it is called, at which point A will be defined.