模板化运算符重载内的成员函数不起作用

Question

我的printTree函数在重载的ostream运算符中不起作用。错误和代码如下。

错误代码：C3861（'printTree'：找不到标识符）

描述：'printTree'：该函数未在模板定义上下文中声明，只能通过以下方式找到实例上下文中与参数相关的查找

类声明

#pragma once
#include "BSTNode.h"
#include <iostream>
template <typename T> class BST;
template <typename T>
std::ostream& operator<<(std::ostream& out, const BST<T>& rhs);

template <typename ItemType>
class BST
{
private: 
    BSTNode<ItemType>* root;
    BSTNode<ItemType>* insert(ItemType* &data, BSTNode<ItemType>* root);
    BSTNode<ItemType>* remove(const ItemType &x, BSTNode<ItemType>* root);
    BSTNode<ItemType>* findMin(BSTNode<ItemType>* root) const;
    BSTNode<ItemType>* findMax(BSTNode<ItemType>* root) const;
    bool search(const ItemType &data, BSTNode<ItemType>* root) const;
    void destroyTree(BSTNode<ItemType>* root); //for destructor
    BSTNode<ItemType>* clone(BSTNode<ItemType>* root) const;
    void printTree(std::ostream& out, BSTNode<ItemType>* root) const;

public:  
    BST();
    ~BST();
    BST(const BST &rhs); 
    BSTNode<ItemType>* getRoot() const {return root;}
    bool isEmpty(); 
    void insert(ItemType* &data);
    bool remove(const ItemType &x);
    ItemType findMin() const; 
    ItemType findMax() const;
    bool search(const ItemType &data) const;
    friend std::ostream& operator<< <>(std::ostream & out ,const BST<ItemType> &rhs);  
    void printTree (std::ostream &out = std::cout) const; 
};

相关方法

template <typename ItemType>
void BST<ItemType>::printTree(std::ostream& out, BSTNode<ItemType>* root) const
{
   if (root != nullptr)
   {
      printTree(out, root->getLeft());
      out << *(root->getData()) << std::endl;
      printTree(out, root->getRight());
   }
}


template <typename ItemType>
std::ostream& operator << <>(std::ostream & out , const BST<ItemType> &rhs)
{
   printTree(out, rhs.getRoot());
   return out; 
}

已解决：

需要在运算符重载定义中将调用对象添加到printTree函数中

template <typename ItemType>
std::ostream& operator << <>(std::ostream & out , const BST<ItemType> &rhs)
{
   rhs.printTree(out, rhs.getRoot());
   return out; 
}

Answer 1

std::ostream& operator<<(std::ostream& out, const BST<T>& rhs);
friend std::ostream& operator<< <>(std::ostream & out ,const BST<ItemType> &rhs);

有区别>，所以我不确定这是否是您的错误。与方法相同。