如何使模板参数的构造函数成为朋友? [英] How do I make a template parameter's constructor a friend?
问题描述
在C ++ 11中，他们使得可以简单地与 friend T 成为模板参数的朋友.您还可以使用 friend T :: Method()将该参数内的方法作为朋友.
In C++11, they made it possible to friend a template parameter simply with friend T. You can also friend methods within that parameter with friend T::Method().
但是，您如何与模板参数的构造函数成为朋友?
However, how do you friend a template parameter's constructor?
class BeMyFriend
{
public:
BeMyFriend& operator=(const BeMyFriend& rhs) = default;
BeMyFriend(const BeMyFriend& rhs) = default;
};
template<class T>
class Test
{
friend T& T::operator=(const T&); //Works fine, no error
friend T::T(const T&); //error: prototype for 'BeMyFriend::BeMyFriend(const BeMyFriend&)' does not match any in class 'BeMyFriend'
};
int main()
{
Test<BeMyFriend> hmm;
return 0;
}
我可以将模板参数的 operator = 很好地作为朋友，但是我不能与 T :: T(const T&)成为朋友.
I'm able to friend the template parameter's operator= just fine, but I'm unable to friend T::T(const T&).
如何使 friend T :: T(const T&); 工作?
这似乎是与要使构造函数成为朋友的问题不同的问题模板类.那里的问题是在声明中处理圆形模板参数.它不处理实际模板参数的构造函数.
edit: This appears to be a different issue than what is solved in Make Friend the constructor of a template class. The issue there is dealing with circular template parameters in a declaration. It doesn't deal with the constructor of an actual template parameter.
Foo 的类型是普通的模板化类，在我的示例中不是像 T 这样的模板参数.该提交中的 friend Foo< B> :: Foo< B>()之类的东西应该可以很好地编译，这与我在 friend T :: T(const T&;).
The type of the Foo is a normal templated class, not a template parameter like T in my example. Something like friend Foo<B>::Foo<B>() from that submission should compile just fine, unlike the issue I'm having here with friend T::T(const T&).
万一这很重要，我将使用gcc 7.2进行编译.
edit: In case this ends up mattering, I'm compiling with gcc 7.2.
我还想澄清一下，C ++确实支持使构造函数成为朋友.例如，第一个示例中的 friend X :: X(char)，X ::〜X(); 位于
edit:
I also want to clarify that C++ does support making constructors friends. For example, friend X::X(char), X::~X(); in the first example at http://en.cppreference.com/w/cpp/language/friend.
这里的问题是如何使模板参数的构造函数成为朋友.
The issue here is how to make a template parameter's constructor a friend.
推荐答案
通过在 friend 之后添加 void ，我可以在GCC中对其进行编译:
I was able to get it to compile in GCC by adding void after friend:
friend void T::T(const T&);
然后，我可以从 BeMyFriend 的构造函数访问 Test 的私有成员之一.但是，请注意，这是编译器特定的.我用clang尝试了一下，但没有起作用.
I was then able to access one of Test's private members from BeMyFriend's constructor. However, note that this is compiler specific. I tried it in clang, and it did not work.
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