在模板类中编写朋友函数声明的正确方法是什么? [英] What is the right way to write friend function declarations in template class?

问题描述

我正在尝试编写自己的矢量模板类，但是在编写朋友函数声明时遇到了一些问题.

I'm trying to write my own vector template class, but I have some problems when writing friend function declarations.

起初我是这样写的:

template <typename T, typename Alloc = std::allocator<T>>
class vector {
public:
    friend bool operator==(const vector<T, Alloc>&, const vector<T, Alloc>&);
};

但是编译器报告警告，我声明了一个非模板函数.因此，我将朋友声明更改为:

But the compiler reports a warning that I declare a non-template function. So I changed the friend declaration to this:

template <typename T, typename Alloc = std::allocator<T>>
class vector {
public:
    template <typename E, typename F>
    friend bool operator==(const vector<E, F>&, const vector<E, F>&);
};

到目前为止，一切都很好，但我认为仍然存在问题.如果我这样写，我将使用所有以两个模板参数作为其朋友功能的operator==函数.例如，operator==(const vector<int>&, const vector<int>&)和operator==(const vector<double>&, const vector<double>&)都将是vector<int>的朋友功能.

So far everything is fine, but I think there are still problems. If I write like that, I make all operator== functions which take two template arguments as its friend functions. For example, operator==(const vector<int>&, const vector<int>&) and operator==(const vector<double>&, const vector<double>&) would both be vector<int>'s friend function.

在模板类中编写朋友函数的正确方法是什么?

What is the right way to write friend functions in template class?

推荐答案

朋友非模板功能

但是编译器报告警告，我声明了一个非模板函数.

But the compiler reports a warning that I declare a non-template function.

是的，您要在类定义中声明一个非模板函数.这意味着，如果您在类定义之外定义它，则必须将其定义为非模板函数，并针对所有可能的实例进行定义，例如:

Yes, you're declaring a non-template function inside the class definition. That means if you define it out of the class definition, you have to define it as non-template function, and for all the possible instantiations , like:

bool operator==(const vector<int>& v1, const vector<int>& v2)
{
    ...
}
bool operator==(const vector<char>& v1, const vector<char>& v2)
{
    ...
}

这很丑陋，您可以在类定义中定义它，例如

That is ugly, you can define it inside the class definition like

template <typename T, typename Alloc = std::allocator<T>>
class vector {
public:
    friend bool operator==(const vector<T, Alloc>&, const vector<T, Alloc>&) {
        ...
    }
};

朋友功能模板

如果要将其定义为模板函数，并限制友谊范围，则可以

Friend function template

If you want to define it as template function, and constrain the scope of friendship, you can

// forward declaration
template <typename T, typename Alloc>
class vector;

// forward declaration
template <typename T, typename Alloc>
bool operator==(const vector<T, Alloc>& v1, const vector<T, Alloc>& v2);

template <typename T, typename Alloc = std::allocator<T>>
class vector {
private:
    int i;
public:
    // only the instantiation of operator== with template parameter type of current T and Alloc becomes friend
    friend bool operator==<>(const vector<T, Alloc>& v1, const vector<T, Alloc>& v2);
};

template <typename T, typename Alloc = std::allocator<T>>
bool operator==(const vector<T, Alloc>& v1, const vector<T, Alloc>& v2)
{
    ...
}

然后，对于vector<int>，只有bool operator==(const vector<int>&, const vector<int>&)是朋友，而其他实例化(例如bool operator==(const vector<double>&, const vector<double>&))则不是.

Then, for vector<int>, only bool operator==(const vector<int>&, const vector<int>&) is friend, other instantiations like bool operator==(const vector<double>&, const vector<double>&) is not.

实时

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