在模板链表中使用友元函数时出现链接错误
我编写了一个模板链表(在 .h 文件中),但出现链接错误.
I programmed a template linked list(in .h file) and I get link error.
template <typename T>
class LinkedList
{
private:
Node<T>* head;
Node<T>* tail;
int size;
public:
LinkedList();
~LinkedList();
inline T* Front() {return &(this->head);};
inline const T* Front() const {return (const T*)this->head;};
void InsertFirst(const T&);
void InsertLast(const T&);
void RemoveFirst();
void RemoveLast ();
void RemoveItem (const T&);
void Sort();
void Clear();
inline bool Exists(const T&) const;
bool Empty() const {return this->size==0 ? true : false;};
inline int Size() const {return this->size;};
T* At(const int index);
const T* At(int index) const;
friend ostream& operator << (ostream& out, const LinkedList<T>& that);
T* operator[](const int);
const T* operator[](const int) const;
};
.
.
.
template <typename T>
ostream& operator << (ostream& out, const LinkedList<T>& that)
{
if (!that.Empty())
for(Node<T>* seeker=that.head; seeker; seeker=seeker->next)
out<<seeker->info<<endl;
return out;
}
由于某种原因,当我在类中的朋友函数的声明中写入时,链接错误消失了:
For some reason the link error disappears when I write instead in the declaration of the friend function in the class:
template <typename T> friend ostream& operator << (ostream& out, const LinkedList<T>& that);
推荐答案
事情是这样的:你声明的朋友不是模板,所以给定的 << 实例化模板不是您宣布为朋友的那个.
Here's the thing: the friend you declared is not a template, so the given instantiation of your << template isn't the one you declared friend.
如果你这样声明朋友
template <typename U> //or T, doesn't matter
friend ostream& operator << (ostream& out, const LinkedList<U>& that);
then operator <<<int>
将成为 LinkedList
的朋友.如果这是不可取的,则有以下解决方案:
then operator << <int>
will be a friend of LinkedList<float>
. If that is undesirable, there is this solution:
friend ostream& operator <<<T> (ostream& out, const LinkedList<T>& that);
在这种情况下,只有模板的特定实例才是您的朋友,这可能正是您所需要的.
In this case, only the particular instantiation of the template is your friend, which might be what you need.
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