我如何调用:: std :: make_shared在一个类只有受保护或私有构造函数？ [英] How do I call ::std::make_shared on a class with only protected or private constructors?

问题描述

我有这个代码不工作，但我想目的是清楚：

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();
}

但我编译时遇到此错误：

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

这个消息基本上是说在模板实例化栈中从 :: std :: make_shared 无法访问构造函数，因为它受到保护。

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.

但我真的想使用 :: std： ：make_shared 并阻止任何人创建一个 :: std :: shared_ptr 指向的类的对象。是否有任何方法可以完成此操作？

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?

推荐答案

这个答案可能更好，可能接受。但我也想出了一个方法，这是更丑陋，但仍然让一切仍然是内联，不需要派生类：

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>

class A {
   struct this_is_private {
       explicit this_is_private(int) {}
   };

 public:
   explicit A(const this_is_private &) {}

   static ::std::shared_ptr<A> create() {
      return ::std::make_shared<A>(this_is_private{0});
   }

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

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

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