問題，我用C++方法

Question

我有一個類，像這樣：

class Qtree 
{ 
public: 
    Qtree(); 
    Qtree(BMP img, int d); 
private: 
    class QtreeNode 
    { 
    public: 
     QtreeNode* nwChild; // pointer to northwest child 
     QtreeNode* neChild; // pointer to northeast child 
     QtreeNode* swChild; // pointer to southwest child 
     QtreeNode* seChild; // pointer to southeast child 
     RGBApixel element; // the pixel stored as this node's "data" 

     QtreeNode(); 
     QuadtreeNode copy(QuadtreeNode & n); 

    }; 

所以問題是在複製方法。它製作給定節點的副本並將其返回。

QtreeNode Qtree::QtreeNode::copy(QtreeNode & n) { 
    QtreeNode *newNode; 
    //.... 
    return newNode; 
} 

然後我打電話副本從我Qtree的拷貝構造函數：

root=QtreeNode::copy(*tree.root); //each tree has a root pointer 
//have also tried many different things here, but dont really know what to put 

我收到以下錯誤：

error: cannot call member function ‘Qtree::QtreeNode* Qtree::QtreeNode::copy(Qtree::QtreeNode&)’ without object 

and 

error: no matching function for call to ‘copy(Qtree::QtreeNode&)’ 

Answer 1

嘗試：

static QuadtreeNode copy(QuadtreeNode & n); 

或更好地創造複製構造函數。

copy是實例方法，它表示它只能用提供的對象運行。它的內部簽名的功能是：

QuadtreeNode copy(QtreeNode* this, QuadtreeNode & n); 

所以你必須傳遞給參數。使用符號obj->copy(..)，但它實際上是通過obj作爲第一個參數。

Answer 2

你需要你的類Qtree :: QtreeNode的一個實例能夠調用它的copy（）方法。

你不能做：

root = Qtree::QtreeNode::copy(*tree.root); 

但你可以這麼做：

Qtree::QtreeNode myQtree; 
root = myQtree.copy(*tree.root); 

Answer 3

QtreeNode Qtree::QtreeNode::copy(QtreeNode & n) { 
    QtreeNode *newNode; 
    //.... 
    return newNode; 
} 

我可能是錯的，但是你定義的函數爲返回（QtreeNode），然後穿上」你返回一個（QtreeNode *）？