如何让 std::make_unique 成为我班的朋友
我想将 std::make_unique
函数声明为我班级的朋友.原因是我想声明我的构造函数 protected
并提供使用 unique_ptr
创建对象的替代方法.这是一个示例代码:
I want to declare std::make_unique
function as a friend of my class. The reason is that I want to declare my constructor protected
and provide an alternative method of creating the object using unique_ptr
. Here is a sample code:
#include <memory>
template <typename T>
class A
{
public:
// Somehow I want to declare make_unique as a friend
friend std::unique_ptr<A<T>> std::make_unique<A<T>>();
static std::unique_ptr<A> CreateA(T x)
{
//return std::unique_ptr<A>(new A(x)); // works
return std::make_unique<A>(x); // doesn't work
}
protected:
A(T x) { (void)x; }
};
int main()
{
std::unique_ptr<A<int>> a = A<int>::CreateA(5);
(void)a;
return 0;
}
现在我收到此错误:
Start
In file included from prog.cc:1:
/usr/local/libcxx-head/include/c++/v1/memory:3152:32: error: calling a protected constructor of class 'A<int>'
return unique_ptr<_Tp>(new _Tp(_VSTD::forward<_Args>(__args)...));
^
prog.cc:13:21: note: in instantiation of function template specialization 'std::__1::make_unique<A<int>, int &>' requested here
return std::make_unique<A>(x); // doesn't work
^
prog.cc:22:41: note: in instantiation of member function 'A<int>::CreateA' requested here
std::unique_ptr<A<int>> a = A<int>::CreateA(5);
^
prog.cc:17:5: note: declared protected here
A(T x) { (void)x; }
^
1 error generated.
1
Finish
将 std::make_unique
声明为我班级朋友的正确方法是什么?
What is the correct way to declare std::make_unique
as a friend of my class?
推荐答案
make_unique
完美转发你传递给它的参数;在您的示例中,您将左值 (x
) 传递给函数,因此它会将参数类型推导出为 int&
.你的 friend
函数声明需要
make_unique
perfect forwards the arguments you pass to it; in your example you're passing an lvalue (x
) to the function, so it'll deduce the argument type as int&
. Your friend
function declaration needs to be
friend std::unique_ptr<A> std::make_unique<A>(T&);
同样,如果你要在 CreateA
中 move(x)
,friend
声明需要是
Similarly, if you were to move(x)
within CreateA
, the friend
declaration would need to be
friend std::unique_ptr<A> std::make_unique<A>(T&&);
这将使代码编译,但绝不保证它会编译另一种实现,因为就您所知,make_unique
将其参数转发给另一个实际实例化您的类的内部辅助函数,在这种情况下,辅助函数需要是 friend
.
This will get the code to compile, but is in no way a guarantee that it'll compile on another implementation because for all you know, make_unique
forwards its arguments to another internal helper function that actually instantiates your class, in which case the helper would need to be a friend
.
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