問題描述

我有一個類模板 Obj 和一個函數模板 make_obj.Obj 定義了一個 private 單個構造函數，該構造函數接受對其要綁定到的模板化類型的引用.

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};
}

我想要的是將 make_obj 函數聲明為 friend 以便它可以創建 Obj 的，但沒有其他人可以(除了通過復制構造函數).

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).

我嘗試了幾個朋友聲明，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&);

后者是對 Obj 類的 make_obj 朋友的所有模板實例化的不太理想的嘗試.但是，在這兩種情況下，我都會遇到相同的錯誤:

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");
             ^

嘗試做 make_obj("hello"); 用于示例目的.

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

如何只允許 make_obj 訪問 Obj 的值構造器?

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

推薦答案

你需要一些前置聲明:

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);
}

實例

順便說一句:我不認為你真的想要 T &t; 用于您的類的成員變量.可能 T t; 是更好的選擇 ;)

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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