我试图重载<<操作符,以便我只能键入cout << linkedList
,但是由于某种原因,我在访问ListType类中的私有NodeType<T> head
时遇到了问题。
重载功能:
template <class U>
std::ostream& operator << (std::ostream& out, const ListType<U>& list) {
if(list.size() > 0) {
NodeType<U>* temp = list.head;
out << temp -> info;
temp = temp -> link;
while(temp != NULL) {
out << ", " << temp -> info;
temp=temp -> link;
}
}
return out;
}
ListType原型:
template <class T>
class ListType {
protected:
NodeType<T>* head;
size_t count;
public:
ListType(); //DONE
ListType(const ListType&); // DONE
virtual ~ListType(); //DONE
const ListType& operator = (const ListType&); //DONE
virtual bool insert(const T&)=0; //DONE
virtual void eraseAll(); //DONE
void erase(const T&); //DONE
bool find(const T&);
size_t size() const; //DONE
bool empty() const;//DONE
private:
void destroy();//DONE
void copy(const ListType&);//DONE
template <class U>
friend std::ostream& operator << (std::ostream&, const ListType&); //DONE
};
NodeType原型:
template <class T>
class NodeType {
public:
T info;
NodeType* link;
};
引发的错误是
NodeType<int>* ListType<int>::head is protected
和
error within this context
答案 0 :(得分:1)
您的friend
声明与operator <<
的声明不匹配。更改
template <class U>
friend std::ostream& operator << (std::ostream&, const ListType&);
到
template <class U>
friend std::ostream& operator << (std::ostream&, const ListType<U>&);
// ^^^