C++繼承和操作符重載

Question

我有兩個類，一個是另一個的父類。他們都需要有不同的操作員功能。每個類都有一個獨特的輸出函數。這裏是他們是如何或多或少設置：

template <class T> 
class Tree { 

    protected: 
     T elem; 
     vector< Tree<T>* > children; 

    public: 
     ostream &output(ostream& stream,int level) { 
      stream << "Tree" << endl; 
      for (int j=0;j<level;j++) stream << " "; 
      stream << '\'' << this->elem << '\'' << endl; 
      for (unsigned int i=0;i<this->children.size();i++) { 
       this->children[i]->output(stream,level+1); 
      } 
      return stream; 
     } 

     template <class U> 
     friend ostream &operator<<(ostream &cout,Tree<U> &obj); 

}; 

template <class T> 
ostream &operator<<(ostream &stream,Tree<T> &obj) { 
    obj.output(stream,0); 
    return stream; 
}; 

---

template <class T> 
class ParseTree : public Tree<T> { 

    protected: 
     ParseTree<T>* elemTree; 
     vector< ParseTree<T>* > children; 

    public: 
     ostream &output(ostream& stream,int level) { 
      stream << "ParseTree" << endl; 
      if (elemTree == NULL) { 
       for (int j=0;j<level;j++) stream << " "; 
       stream << '\'' << this->elem << '\'' << endl; 
      } 
      else { 
       elemTree->output(stream,level+1); 
      } 
      for (unsigned int i=0;i<this->children.size();i++) { 
       this->children[i]->output(stream,level+1); 
      } 
      return stream; 
     } 

     template <class U> 
     friend ostream &operator<<(ostream &cout,ParseTree<U> &obj); 
}; 

template <class T> 
ostream &operator<<(ostream &stream,ParseTree<T> &obj) { 
    obj.output(stream,0); 
    return stream; 
}; 

兩個輸出功能遞歸地打印出樹，但分析樹的略有不同。我遇到的問題是，當我嘗試輸出一個ParseTree時，第一次迭代來自ParseTree的輸出函數（通過流< <「ParseTree」< < endl語句確認），但所有後續請求似乎都是Tree輸出功能（如由流確認的< <「樹」< < endl語句）。每一個推到子載體上的對象都是ParseTree。我的猜測是ParseTree :: children與Tree :: children不同，並且出於某種原因上下文正在切換。有任何想法嗎？

Answer 1

使輸出virtual和定義operator<<只有一次與Tree<T>類型的參數。這應該足夠了。