c++ - 如何为嵌套在一个模板类里面的类添加友元？比如==函数。

Question

template <typename> class Vector;
template <typename T>
bool operator==(const typename Vector<T>::const_iterator&, const typename Vector<T>::const_iterator&); // 友元声明，签名是这么写吗？

template <typename T>
class Vector {
public:
    class const_iterator: {
        friend bool operator==(const const_iterator& lsh, const const_iteraotr& rhs);
    public:
    };
};


template <typename T>
bool operator==(const typename Vector<T>::const_iterator& lhs, const typename Vector<T>::const_iterator& rhs) { // 友元定义

}

如果直接将==友元在const_iterator里面定义，比较简单，直接写就行。但是我想在类外定义时，就不知道怎么写它的函数签名了。

Answer 1

It is best to write it like this, what you wrote above seems difficult to achieve
Reference

template <typename> struct Vector;

template <typename T>
struct Vector_const_iterator;

template <typename T>
bool operator==(const Vector_const_iterator<T>& lsh, const Vector_const_iterator<T>& rhs);

template <typename T>
struct Vector {
    typedef Vector_const_iterator<T> const_iterator;
};

template <typename T>
struct Vector_const_iterator {
    friend bool operator == <> (const Vector_const_iterator<T>& lsh, const Vector_const_iterator<T>& rhs);
};


template <typename T>
bool operator==(const Vector_const_iterator<T>& lsh, const Vector_const_iterator<T>& rhs) { 
    return true;
}

Answer 2

There is something wrong with your function declaration. Friends only have access rights, not members of the class.

template <typename T>
class Vector {
public:
    class const_iterator{
        friend bool operator==(const typename Vector<T>::const_iterator& lsh, 
                               const typename Vector<T>::const_iterator& rhs);
    public:
    };
};


template <typename T>
bool operator==(const typename Vector<T>::const_iterator& lhs, const typename Vector<T>::const_iterator& rhs) { // 友元定义
  return false;
}