C++ 声明模板类的模板友元函数
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declare template friend function of template class
提问by Ryan Haining
I have a class template Objand a function template make_obj. Objhas a privatesingle constructor defined, which takes a reference to its templated type to bind to.
我有一个类模板Obj和一个函数模板make_obj。 定义Obj了private一个构造函数，该构造函数接受对其要绑定到的模板化类型的引用。
template <typename T>
class Obj {
private:
T& t;
Obj(T& t)
: t{t}
{ }
};
template <typename T>
Obj<T> make_obj(T& t) {
return {t};
}
What I want is to declare the make_objfunction a friendso that it can create Obj's, but no one else can (except via the copy ctor).
我想要的是声明make_obj函数 afriend以便它可以创建Obj's，但没有其他人可以（通过复制 ctor 除外）。
I have tried several friend declaration including
我尝试了几个朋友声明，包括
friend Obj make_obj(T&);
and
和
template <typename T1, typename T2>
friend Obj<T1> make_obj(T2&);
The latter being a less than desirable attempt at making all template instantiations of make_objfriends of the Objclass. However in both of these cases I get the same error:
后者是make_obj对Obj类的朋友的所有模板实例化的不太理想的尝试。但是，在这两种情况下，我都会遇到相同的错误：
error: calling a private constructor of class 'Obj<char const[6]>'
return {t};
^
note: in instantiation of function template specialization
'make_obj<const char *>' requested here
auto s = make_obj("hello");
^
trying to do make_obj("hello");for example purposes.
试图做make_obj("hello");的例子目的。
How can I allow only make_objaccess to Obj's value contructor?
如何只允许make_obj访问Obj的值构造函数？
回答by Daniel Frey
You need a few forward declarations:
您需要一些前向声明：
template <typename T>
class Obj;
template <typename T>
Obj<T> make_obj(T t);
template <typename T>
class Obj {
private:
T & t;
Obj (T & t) : t(t) { }
Obj() = delete;
friend Obj make_obj<T>(T t);
};
template <typename T>
Obj<T> make_obj(T t) {
return Obj<T>(t);
}
And BTW: I don't think you really want T & t;for your class' member variable. Probably T t;is a better choice ;)
顺便说一句：我不认为你真的想要T & t;你班级的成员变量。可能T t;是更好的选择;)
回答by Curious
With the automatic return type syntax, you only need to forward declare the function and everything works. Here is an example
使用自动返回类型语法，您只需要向前声明函数，一切正常。这是一个例子
template <typename T>
auto make_obj(T t);
template <typename T>
class Obj {
private:
T & t;
Obj (T & t) : t(t) { }
Obj() = delete;
friend auto make_obj<T>(T t);
};
template <typename T>
auto make_obj(T t) {
return Obj<T>{t};
}
int main() {
make_obj(1);
return 0;
}
