如何使模板成为T的朋友? [英] How to make a template a friend of T?

问题描述

假设我有一个C类和一个容器模板T_CONTAINER.
我有一个日期函数，只想向T_CONTAINER公开.
我怎样才能做到这一点?请注意，我使用Visual C ++ 2010作为编译器.

我不得不公开这些功能，这是不理想的.

我尝试过:

Let''s say I have a class C and a container template T_CONTAINER.
I have a function in date that I only want to expose to T_CONTAINER.
How can I do this? Note that I am using Visual C++ 2010 as a compiler.

I''ve had to make the functions public, which is less than desirable.

I''ve tried:

class C;
#include "T_CONTAINER.H"
class C {
     friend T_CONTAINER< C >;
}

推荐答案

尝试了以下可行的方法.

Tried the following which works.

template <typename T>
class A
{
    T *_t;
public:
    A(T *t) : _t(t) {}
    void accessT()
    {
        _t->privateAccess();
    }
};
class C {
    void privateAccess()
    {
        cout << "private" << endl;
    }
    friend A<C>;
};
int main()
{
    C c;
    A<C> a(&c);
    a.accessT();
    return 0;
}

使用friend声明，模板类可以访问类C的私有方法.我不知道您在哪里出错.

大多数容器类都有一些用于分配器等的默认模板参数.确保它们在您的朋友声明中匹配.

With the friend declaration, the template class can access the private methods of class C. I don''t know where you went wrong.

Most container classes have some default template arguments for allocators and such. Make sure they match in your friend declaration.

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