C ++模板:部分模板规范和朋友类 [英] C++ Templates: Partial Template Specifications and Friend Classes

问题描述

是否有可能使部分模板规范成为朋友类? IE.认为您具有以下模板类

is it possible to somehow make a partial template specification a friend class? I.e. consider you have the following template class

template <class T> class X{
    T t;
};

现在您有部分专长，例如，指针

Now you have partial specializations, for example, for pointers

template <class T> class X<T*>{
    T* t;
};

我想完成的是，每个可能的X<T*>是X<S>的任何S的朋友类. IE. X<A*>应该是X<B>的朋友.

What I want to accomplish is that every possible X<T*> is a friend class of X<S> for ANY S. I.e. X<A*> should be a friend of X<B>.

当然，我想到了X中常用的模板朋友声明:

Of course, I thought about a usual template friend declaration in X:

template <class T> class X{
    template <class S> friend class X<S*>;
}

但是，这无法编译，g ++告诉我:

However, this does not compile, g++ tells me this:

test4.cpp:34:15:错误:"template<class T> class X"的专业化必须出现在名称空间范围内

test4.cpp:34:15: error: specialization of 'template<class T> class X' must appear at namespace scope

test4.cpp:34:21:错误:部分专业化'X<S*>'被声明为'朋友'

test4.cpp:34:21: error: partial specialization 'X<S*>' declared 'friend'

这根本不可能还是有解决方法?

Is this not possible at all or is there some workaround?

我问的原因是，我需要一个X<T*>中的构造函数，该构造函数可以从任意X<S>(S必须是T的子类型)创建此类.

The reason why I am asking is that I need a constructor in X<T*> that creates this class from an arbitrary X<S> (S must be a subtype of T).

代码如下:

template <class T> class X<T*>{
    T* t;

    template<class S>
    X(X<S> x) : t(&(x.t))  {} //Error, x.t is private
}

现在，编译器当然会抱怨x.t在构造函数中不可见，因为它是私有的.这就是为什么我需要部分专业化朋友课程的原因.

Now, the compiler complains, of course, that x.t is not visibile in the constructor since it is private. This is why I need a partial specialization friend class.

推荐答案

在C ++中，您可以在四个级别上授予private以外的权限.

In C++, you can grant access beyond private on four levels.

• 完全public访问(请参阅pmr的答案)
• 继承层次结构内的访问权限(protected，此处无关)
• 到基本模板friend(请参阅此答案)
• 非模板或完全专用的friend(太弱而无法解决您的用例)

• completely public access (see pmr's answer)
• access within inheritance hierarchy (protected, irrelevant here)
• to a base template friend (see this answer)
• to a non-template or fully specialized friend (too weak to solve your use case)

后两种友谊之间没有中间途径.

摘自C ++标准的第14.5.4节:.

From §14.5.4 of the C++ standard:.

朋友声明不能声明部分专业化.

Friend declarations shall not declare partial specializations.

以下声明将使您能够实现所需的内容.它使您可以自由地访问模板的任何其他专业化内容，但仍然只能在X中进行.它比您要求的要宽松一些.

The following declaration will allow you to implement what you need. It gives you a free hand to access any specialization of your template from any other specialization, but still only within X. It is slightly more permissive than what you asked for.

template<class T> class X
{
    template<class Any> friend class X;
    public:
        ...
};

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