声明模板类的模板友元函数

问题描述

我有一个类模板 Obj 和一个函数模板 make_obj.Obj 定义了一个 private 单个构造函数，该构造函数接受对其要绑定到的模板化类型的引用.

I have a class template Obj and a function template make_obj. Obj has a private single constructor defined, which takes a reference to its templated type to bind to.

template <typename T>
class Obj {
  private:
    T& t;
    Obj(T& t)
        : t{t}
    { }
};

template <typename T>
Obj<T> make_obj(T& t) { 
    return {t};
}

我想要的是将 make_obj 函数声明为 friend 以便它可以创建 Obj 的，但没有其他人可以(除了通过复制构造函数).

What I want is to declare the make_obj function a friend so that it can create Obj's, but no one else can (except via the copy ctor).

我尝试了几个朋友声明，包括

I have tried several friend declaration including

friend Obj make_obj(T&);

和

template <typename T1, typename T2>
friend Obj<T1> make_obj(T2&);

后者是对 Obj 类的 make_obj 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:

The latter being a less than desirable attempt at making all template instantiations of make_obj friends of the Obj class. However in both of these cases I get the same error:

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");
             ^

尝试做 make_obj("hello"); 用于示例目的.

trying to do make_obj("hello"); for example purposes.

如何只允许 make_obj 访问 Obj 的值构造器?

How can I allow only make_obj access to Obj's value contructor?

推荐答案

你需要一些前置声明:

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);
}

实例

顺便说一句:我不认为你真的想要 T &t; 用于您的类的成员变量.可能 T t; 是更好的选择 ;)

And BTW: I don't think you really want T & t; for your class' member variable. Probably T t; is a better choice ;)

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