模板类的模板朋友功能 [英] template friend functions of template class

问题描述

我具有以下模板类和模板函数，旨在访问该类的私有数据成员:

I have the following template class and template function which intends to access the class' private data member:

#include <iostream>

template<class T>
class MyVar
{
    int x;
};

template<class T>
void printVar(const MyVar<T>& var)
{
    std::cout << var.x << std::endl;
}

template<class T>
void scanVar(MyVar<T>& var)
{
    std::cin >> var.x;
}

struct Foo {};

int main(void)
{
    MyVar<Foo> a;
    scanVar(a);
    printVar(a);
    return 0;
}

要将两个函数声明为MyVar<T>的朋友函数，我尝试了在template<class T> class MyVar声明中声明友谊的以下方法.它们都不起作用.我该怎么办?

To declare the two functions as MyVar<T>'s friend functions, I've tried the following ways inside the declaration of template<class T> class MyVar to declare friendship. None of them works. How should I do?

template<class T> friend void printVar(const MyVar&);
template<class T> friend void scanVar(MyVar&);
// compilation error

template<class T> friend void printVar(const MyVar<T>&);
template<class T> friend void scanVar(MyVar<T>&);
// compilation error

friend void printVar(const MyVar<T>&);
friend void scanVar(MyVar<T>&);
// link error

friend void printVar(const MyVar&);
friend void scanVar(MyVar&);
// link error too

推荐答案

最简单的选择是在类中定义朋友:

The simplest option is to define the friend within the class:

template<class T>
class MyVar
{
    int x;

    friend void printVar(const MyVar & var) {
        std::cout << var.x << std::endl;
    }
    friend void scanVar(MyVar & var) {
        std::cin >> var.x;
    }
};

缺点是只能通过依赖于参数的查找来调用函数.在您的示例中这不是问题，但是如果他们没有合适的参数，或者您想在不调用名称的情况下指定名称，则可能是一个问题.

The downside is that the functions can only be called through argument-dependent lookup. That's not a problem in your example, but might be a problem if they don't have a suitable argument, or you want to specify the name without calling it.

如果要使用单独的定义，则必须在类定义之前声明模板(以便可用于朋友声明)，然后在之后定义(以便它可以访问类成员).该类也必须在函数之前声明.这有点混乱，所以我只显示两个功能之一:

If you want a separate definition, then the template will have to be declared before the class definition (so it's available for a friend declaration), but defined afterwards (so it can access the class members). The class will also have to be declared before the function. This is a bit messy, so I'll only show one of the two functions:

template <typename T> class MyVar;
template <typename T> void printVar(const MyVar<T> & var);

template<class T>
class MyVar
{
    int x;

    friend void printVar<T>(const MyVar<T> & var);
};

template <typename T> void printVar(const MyVar<T> & var) {
    std::cout << var.x << std::endl;
}

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