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

{代码...} 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。

高洛峰 · Answer

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

template  struct Vector;

template 
struct Vector_const_iterator;

template 
bool operator==(const Vector_const_iterator& lsh, const Vector_const_iterator& rhs);

template 
struct Vector {
    typedef Vector_const_iterator const_iterator;
};

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


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

PHP中文网 · Answer

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

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


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