How do I call ::std::make_shared on a class with only protected or private constructors?

Question

I have this code that doesn't work, but I think the intent is clear:

testmakeshared.cpp

#include <memory>

class A {
 public:
   static ::std::shared_ptr<A> create() {
      return ::std::make_shared<A>();
   }

 protected:
   A() {}
   A(const A &) = delete;
   const A &operator =(const A &) = delete;
};

::std::shared_ptr<A> foo()
{
   return A::create();
}

But I get this error when I compile it:

g++ -std=c++0x -march=native -mtune=native -O3 -Wall testmakeshared.cpp
In file included from /usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:52:0,
                 from /usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/memory:86,
                 from testmakeshared.cpp:1:
testmakeshared.cpp: In constructor ‘std::_Sp_counted_ptr_inplace<_Tp, _Alloc, _Lp>::_Sp_counted_ptr_inplace(_Alloc) [with _Tp = A, _Alloc = std::allocator<A>, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’:
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:518:8:   instantiated from ‘std::__shared_count<_Lp>::__shared_count(std::_Sp_make_shared_tag, _Tp*, const _Alloc&, _Args&& ...) [with _Tp = A, _Alloc = std::allocator<A>, _Args = {}, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:986:35:   instantiated from ‘std::__shared_ptr<_Tp, _Lp>::__shared_ptr(std::_Sp_make_shared_tag, const _Alloc&, _Args&& ...) [with _Alloc = std::allocator<A>, _Args = {}, _Tp = A, __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:313:64:   instantiated from ‘std::shared_ptr<_Tp>::shared_ptr(std::_Sp_make_shared_tag, const _Alloc&, _Args&& ...) [with _Alloc = std::allocator<A>, _Args = {}, _Tp = A]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:531:39:   instantiated from ‘std::shared_ptr<_Tp> std::allocate_shared(const _Alloc&, _Args&& ...) [with _Tp = A, _Alloc = std::allocator<A>, _Args = {}]’
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr.h:547:42:   instantiated from ‘std::shared_ptr<_Tp1> std::make_shared(_Args&& ...) [with _Tp = A, _Args = {}]’
testmakeshared.cpp:6:40:   instantiated from here
testmakeshared.cpp:10:8: error: ‘A::A()’ is protected
/usr/lib/gcc/x86_64-redhat-linux/4.6.1/../../../../include/c++/4.6.1/bits/shared_ptr_base.h:400:2: error: within this context

Compilation exited abnormally with code 1 at Tue Nov 15 07:32:58

This message is basically saying that some random method way down in the template instantiation stack from ::std::make_shared can't access the constructor because it's protected.

But I really want to use both ::std::make_shared and prevent anybody from making an object of this class that isn't pointed at by a ::std::shared_ptr. Is there any way to accomplish this?

Answer 1

This answer is probably better, and the one I'll likely accept. But I also came up with a method that's uglier, but does still let everything still be inline and doesn't require a derived class:

#include <memory>
#include <string>

class A {
 protected:
   struct this_is_private;

 public:
   explicit A(const this_is_private &) {}
   A(const this_is_private &, ::std::string, int) {}

   template <typename... T>
   static ::std::shared_ptr<A> create(T &&...args) {
      return ::std::make_shared<A>(this_is_private{0},
                                   ::std::forward<T>(args)...);
   }

 protected:
   struct this_is_private {
       explicit this_is_private(int) {}
   };

   A(const A &) = delete;
   const A &operator =(const A &) = delete;
};

::std::shared_ptr<A> foo()
{
   return A::create();
}

::std::shared_ptr<A> bar()
{
   return A::create("George", 5);
}

::std::shared_ptr<A> errors()
{
   ::std::shared_ptr<A> retval;

   // Each of these assignments to retval properly generates errors.
   retval = A::create("George");
   retval = new A(A::this_is_private{0});
   return ::std::move(retval);
}

Edit 2017-01-06: I changed this to make it clear that this idea is clearly and simply extensible to constructors that take arguments because other people were providing answers along those lines and seemed confused about this.

Answer 2

Looking at the requirements for std::make_shared in 20.7.2.2.6 shared_ptr creation [util.smartptr.shared.create], paragraph 1:

Requires: The expression ::new (pv) T(std::forward<Args>(args)...), where pv has type void* and points to storage suitable to hold an object of type T, shall be well formed. A shall be an allocator (17.6.3.5). The copy constructor and destructor of A shall not throw exceptions.

Since the requirement is unconditionally specified in terms of that expression and things like scope aren't taken into account, I think tricks like friendship are right out.

A simple solution is to derive from A. This needn't require making A an interface or even a polymorphic type.

// interface in header
std::shared_ptr<A> make_a();

// implementation in source
namespace {

struct concrete_A: public A {};

} // namespace

std::shared_ptr<A>
make_a()
{
    return std::make_shared<concrete_A>();
}

Answer 3

Possibly the simplest solution. Based on the previous answer by Mohit Aron and incorporating dlf's suggestion.

#include <memory>

class A
{
public:
    static std::shared_ptr<A> create()
    {
        struct make_shared_enabler : public A {};

        return std::make_shared<make_shared_enabler>();
    }

private:
    A() {}  
};

Answer 4

Here's a neat solution for this:

#include <memory>

class A {
   public:
     static shared_ptr<A> Create();

   private:
     A() {}

     struct MakeSharedEnabler;   
 };

struct A::MakeSharedEnabler : public A {
    MakeSharedEnabler() : A() {
    }
};

shared_ptr<A> A::Create() {
    return make_shared<MakeSharedEnabler>();
}

Answer 5

How about this?

static std::shared_ptr<A> create()
{
    std::shared_ptr<A> pA(new A());
    return pA;
}

Answer 6

struct A {
public:
  template<typename ...Arg> std::shared_ptr<A> static create(Arg&&...arg) {
    struct EnableMakeShared : public A {
      EnableMakeShared(Arg&&...arg) :A(std::forward<Arg>(arg)...) {}
    };
    return std::make_shared<EnableMakeShared>(std::forward<Arg>(arg)...);
  }
  void dump() const {
    std::cout << a_ << std::endl;
  }
private:
  A(int a) : a_(a) {}
  A(int i, int j) : a_(i + j) {}
  A(std::string const& a) : a_(a.size()) {}
  int a_;
};

Answer 7

Since I didn't like the already provided answers I decided to search on and found a solution that is not as generic as the previous answers but I like it better(tm). In retrospect it is not much nicer than the one provided by Omnifarius but there could be other people who like it too :)

This is not invented by me, but it is the idea of Jonathan Wakely (GCC developer).

Unfortunately it does not work with all the compilers because it relies on a small change in std::allocate_shared implementation. But this change is now a proposed update for the standard libraries, so it might get supported by all the compilers in the future. It works on GCC 4.7.

C++ standard Library Working Group change request is here: http://lwg.github.com/issues/lwg-active.html#2070

The GCC patch with an example usage is here: http://old.nabble.com/Re%3A--v3--Implement-pointer_traits-and-allocator_traits-p31723738.html

The solution works on the idea to use std::allocate_shared (instead of std::make_shared) with a custom allocator that is declared friend to the class with the private constructor.

The example from the OP would look like this:

#include <memory>

template<typename Private>
struct MyAlloc : std::allocator<Private>
{
    void construct(void* p) { ::new(p) Private(); }
};

class A {
    public:
        static ::std::shared_ptr<A> create() {
            return ::std::allocate_shared<A>(MyAlloc<A>());
        }

    protected:
        A() {}
        A(const A &) = delete;
        const A &operator =(const A &) = delete;

        friend struct MyAlloc<A>;
};

int main() {
    auto p = A::create();
    return 0;
}

A more complex example that is based on the utility I'm working on. With this I could not use Luc's solution. But the one by Omnifarius could be adapted. Not that while in the previous example everybody can create an A object using the MyAlloc in this one there is not way to create A or B besides the create() method.

#include <memory>

template<typename T>
class safe_enable_shared_from_this : public std::enable_shared_from_this<T>
{
    public:
    template<typename... _Args>
        static ::std::shared_ptr<T> create(_Args&&... p_args) {
            return ::std::allocate_shared<T>(Alloc(), std::forward<_Args>(p_args)...);
        }

    protected:
    struct Alloc : std::allocator<T>
    {  
        template<typename _Up, typename... _Args>
        void construct(_Up* __p, _Args&&... __args)
        { ::new((void *)__p) _Up(std::forward<_Args>(__args)...); }
    };
    safe_enable_shared_from_this(const safe_enable_shared_from_this&) = delete;
    safe_enable_shared_from_this& operator=(const safe_enable_shared_from_this&) = delete;
};

class A : public safe_enable_shared_from_this<A> {
    private:
        A() {}
        friend struct safe_enable_shared_from_this<A>::Alloc;
};

class B : public safe_enable_shared_from_this<B> {
    private:
        B(int v) {}
        friend struct safe_enable_shared_from_this<B>::Alloc;
};

int main() {
    auto a = A::create();
    auto b = B::create(5);
    return 0;
}

Answer 8

Ideally, I think the perfect solution would require additions to the C++ standard. Andrew Schepler proposes the following:

(Go here for the whole thread)

we can borrow an idea from boost::iterator_core_access. I propose a new class std::shared_ptr_access with no public or protected members, and to specify that for std::make_shared(args...) and std::alloc_shared(a, args...), the expressions ::new(pv) T(forward(args)...) and ptr->~T() must be well-formed in the context of std::shared_ptr_access.

An implementation of std::shared_ptr_access might look like:

namespace std {
    class shared_ptr_access
    {
        template <typename _T, typename ... _Args>
        static _T* __construct(void* __pv, _Args&& ... __args)
        { return ::new(__pv) _T(forward<_Args>(__args)...); }

        template <typename _T>
        static void __destroy(_T* __ptr) { __ptr->~_T(); }

        template <typename _T, typename _A>
        friend class __shared_ptr_storage;
    };
}

Usage

If/when the above is added to the standard, we would simply do:

class A {
public:
   static std::shared_ptr<A> create() {
      return std::make_shared<A>();
   }

 protected:
   friend class std::shared_ptr_access;
   A() {}
   A(const A &) = delete;
   const A &operator =(const A &) = delete;
};

If this also sounds like an important addition to the standard to you, feel free to add your 2 cents to the linked isocpp Google Group.

Answer 9

I realise this thread is rather old, but I found an answer that does not require inheritance or extra arguments to the constructor that I couldn't see elsewhere. It is not portable though:

#include <memory>

#if defined(__cplusplus) && __cplusplus >= 201103L
#define ALLOW_MAKE_SHARED(x) friend void __gnu_cxx::new_allocator<test>::construct<test>(test*);
#elif defined(_WIN32) || defined(WIN32)
#if defined(_MSC_VER) && _MSC_VER >= 1800
#define ALLOW_MAKE_SHARED(x) friend class std::_Ref_count_obj;
#else
#error msc version does not suport c++11
#endif
#else
#error implement for platform
#endif

class test {
    test() {}
    ALLOW_MAKE_SHARED(test);
public:
    static std::shared_ptr<test> create() { return std::make_shared<test>(); }

};
int main() {
    std::shared_ptr<test> t(test::create());
}

I have tested on windows and linux, it may need tweaking for different platforms.

Answer 10

If you also want to enable a constuctor that takes arguments, this may help a bit.

#include <memory>
#include <utility>

template<typename S>
struct enable_make : public S
{
    template<typename... T>
    enable_make(T&&... t)
        : S(std::forward<T>(t)...)
    {
    }
};

class foo
{
public:
    static std::unique_ptr<foo> create(std::unique_ptr<int> u, char const* s)
    {
        return std::make_unique<enable_make<foo>>(std::move(u), s);
    }
protected:
    foo(std::unique_ptr<int> u, char const* s)
    {
    }
};

void test()
{
    auto fp = foo::create(std::make_unique<int>(3), "asdf");
}

Answer 11

I had the same problem, but none of the existing answers were really satisfactory as I need to pass arguments to the protected constructor. Moreover, I need to do this for several classes, each taking different arguments.

To that effect, and building on several of the existing answers which all use similar methods, I present this little nugget:

template < typename Object, typename... Args >
inline std::shared_ptr< Object >
protected_make_shared( Args&&... args )
{
  struct helper : public Object
  {
    helper( Args&&... args )
      : Object{ std::forward< Args >( args )... }
    {}
  };

  return std::make_shared< helper >( std::forward< Args >( args )... );
}

Answer 12

There's a more hairy and interesting problem that happens when you have two strictly related classes A and B that work together.

Say A is the "master class" and B its "slave". If you want to restrict instantiation of B only to A, you'd make B's constructor private, and friend B to A like this

class B
{
public:
    // B your methods...

private:
    B();
    friend class A;
};

Unfortunately calling std::make_shared<B>() from a method of A will make the compiler complain about B::B() being private.

My solution to this is to create a public Pass dummy class (just like nullptr_t) inside B that has private constructor and is friend with A and make B's constructor public and add Pass to its arguments, like this.

class B
{
public:
  class Pass
  {
    Pass() {}
    friend class A;
  };

  B(Pass, int someArgument)
  {
  }
};

class A
{
public:
  A()
  {
    // This is valid
    auto ptr = std::make_shared<B>(B::Pass(), 42);
  }
};

class C
{
public:
  C()
  {
    // This is not
    auto ptr = std::make_shared<B>(B::Pass(), 42);
  }
};

Answer 13

[Edit] I read through the thread noted above on a standardized std::shared_ptr_access<> proposal. Within there was a response noting a fix to std::allocate_shared<> and an example of its use. I've adapted it to a factory template below, and tested it under gcc C++11/14/17. It works with std::enable_shared_from_this<> as well, so would obviously be preferable to my original solution in this answer. Here it is...

#include <iostream>
#include <memory>

class Factory final {
public:
    template<typename T, typename... A>
    static std::shared_ptr<T> make_shared(A&&... args) {
        return std::allocate_shared<T>(Alloc<T>(), std::forward<A>(args)...);
    }
private:
    template<typename T>
    struct Alloc : std::allocator<T> {
        template<typename U, typename... A>
        void construct(U* ptr, A&&... args) {
            new(ptr) U(std::forward<A>(args)...);
        }
        template<typename U>
        void destroy(U* ptr) {
            ptr->~U();
        }
    };  
};

class X final : public std::enable_shared_from_this<X> {
    friend class Factory;
private:
    X()      { std::cout << "X() addr=" << this << "\n"; }
    X(int i) { std::cout << "X(int) addr=" << this << " i=" << i << "\n"; }
    ~X()     { std::cout << "~X()\n"; }
};

int main() {
    auto p1 = Factory::make_shared<X>(42);
    auto p2 = p1->shared_from_this();
    std::cout << "p1=" << p1 << "\n"
              << "p2=" << p2 << "\n"
              << "count=" << p1.use_count() << "\n";
}

---

[Orig] I found a solution using the shared pointer aliasing constructor. It allows both the ctor and dtor to be private, as well as use of the final specifier.

#include <iostream>
#include <memory>

class Factory final {
public:
    template<typename T, typename... A>
    static std::shared_ptr<T> make_shared(A&&... args) {
        auto ptr = std::make_shared<Type<T>>(std::forward<A>(args)...);
        return std::shared_ptr<T>(ptr, &ptr->type);
    }
private:
    template<typename T>
    struct Type final {
        template<typename... A>
        Type(A&&... args) : type(std::forward<A>(args)...) { std::cout << "Type(...) addr=" << this << "\n"; }
        ~Type() { std::cout << "~Type()\n"; }
        T type;
    };
};

class X final {
    friend struct Factory::Type<X>;  // factory access
private:
    X()      { std::cout << "X() addr=" << this << "\n"; }
    X(int i) { std::cout << "X(...) addr=" << this << " i=" << i << "\n"; }
    ~X()     { std::cout << "~X()\n"; }
};

int main() {
    auto ptr1 = Factory::make_shared<X>();
    auto ptr2 = Factory::make_shared<X>(42);
}

Note that the approach above doesn't play well with std::enable_shared_from_this<> because the initial std::shared_ptr<> is to the wrapper and not the type itself. We can address this with an equivalent class that is compatible with the factory...

#include <iostream>
#include <memory>

template<typename T>
class EnableShared {
    friend class Factory;  // factory access
public:
    std::shared_ptr<T> shared_from_this() { return weak.lock(); }
protected:
    EnableShared() = default;
    virtual ~EnableShared() = default;
    EnableShared<T>& operator=(const EnableShared<T>&) { return *this; }  // no slicing
private:
    std::weak_ptr<T> weak;
};

class Factory final {
public:
    template<typename T, typename... A>
    static std::shared_ptr<T> make_shared(A&&... args) {
        auto ptr = std::make_shared<Type<T>>(std::forward<A>(args)...);
        auto alt = std::shared_ptr<T>(ptr, &ptr->type);
        assign(std::is_base_of<EnableShared<T>, T>(), alt);
        return alt;
    }
private:
    template<typename T>
    struct Type final {
        template<typename... A>
        Type(A&&... args) : type(std::forward<A>(args)...) { std::cout << "Type(...) addr=" << this << "\n"; }
        ~Type() { std::cout << "~Type()\n"; }
        T type;
    };
    template<typename T>
    static void assign(std::true_type, const std::shared_ptr<T>& ptr) {
        ptr->weak = ptr;
    }
    template<typename T>
    static void assign(std::false_type, const std::shared_ptr<T>&) {}
};

class X final : public EnableShared<X> {
    friend struct Factory::Type<X>;  // factory access
private:
    X()      { std::cout << "X() addr=" << this << "\n"; }
    X(int i) { std::cout << "X(...) addr=" << this << " i=" << i << "\n"; }
    ~X()     { std::cout << "~X()\n"; }
};

int main() {
    auto ptr1 = Factory::make_shared<X>();
    auto ptr2 = ptr1->shared_from_this();
    std::cout << "ptr1=" << ptr1.get() << "\nptr2=" << ptr2.get() << "\n";
}

Lastly, somebody said clang complained about Factory::Type being private when used as a friend, so just make it public if that's the case. Exposing it does no harm.

Answer 14

The root of the problem is that if the function or class you friend makes lower level calls to your constructor, they have to be friended too. std::make_shared isn't the function that's actually calling your constructor so friending it makes no difference.

class A;
typedef std::shared_ptr<A> APtr;
class A
{
    template<class T>
    friend class std::_Ref_count_obj;
public:
    APtr create()
    {
        return std::make_shared<A>();
    }
private:
    A()
    {}
};

std::_Ref_count_obj is actually calling your constructor, so it needs to be a friend. Since that's a bit obscure, I use a macro

#define SHARED_PTR_DECL(T) \
class T; \
typedef std::shared_ptr<T> ##T##Ptr;

#define FRIEND_STD_MAKE_SHARED \
template<class T> \
friend class std::_Ref_count_obj;

Then your class declaration looks fairly simple. You can make a single macro for declaring the ptr and the class if you prefer.

SHARED_PTR_DECL(B);
class B
{
    FRIEND_STD_MAKE_SHARED
public:
    BPtr create()
    {
        return std::make_shared<B>();
    }
private:
    B()
    {}
};

This is actually an important issue. To make maintainable, portable code you need to hide as much of the implementation as possible.

typedef std::shared_ptr<A> APtr;

hides how you're handling your smart pointer a bit, you have to be sure to use your typedef. But if you always have to create one using make_shared, it defeats the purpose.

The above example forces code using your class to use your smart pointer constructor, which means that if you switch to a new flavor of smart pointer, you change your class declaration and you have a decent chance of being finished. DO NOT assume your next boss or project will use stl, boost etc. plan for changing it someday.

Doing this for almost 30 years, I've paid a big price in time, pain and side effects to repair this when it was done wrong years ago.

Answer 15

#include <iostream>
#include <memory>

class A : public std::enable_shared_from_this<A>
{
private:
    A(){}
    explicit A(int a):m_a(a){}
public:
    template <typename... Args>
    static std::shared_ptr<A> create(Args &&... args)
    {
        class make_shared_enabler : public A
        {
        public:
            make_shared_enabler(Args &&... args):A(std::forward<Args>(args)...){}
        };
        return std::make_shared<make_shared_enabler>(std::forward<Args>(args)...);
    }

    int val() const
    {
        return m_a;
    }
private:
    int m_a=0;
};

int main(int, char **)
{
    std::shared_ptr<A> a0=A::create();
    std::shared_ptr<A> a1=A::create(10);
    std::cout << a0->val() << " " << a1->val() << std::endl;
    return 0;
}