Question

我有一个类似的课程：

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);

   };

因此问题在于复制方法。它生成给定节点的副本并返回它。

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

然后我从我的Qtree拷贝构造函数中调用copy：

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

我收到以下错误：

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

尝试：

static QuadtreeNode copy(QuadtreeNode & n);

或更好地创建复制构造函数。

copy是实例方法，它意味着它只能在提供对象的情况下运行。它的内部签名是一个函数：

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

所以你必须传递给参数。使用了表示法obj->copy(..)，但它实际上是将obj作为第一个参数传递。

Answer 2

您需要一个类Qtree :: QtreeNode的实例才能在其上调用copy（）方法。

你做不到：

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

但你可以这样做：

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

Answer 3

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

我可能错了，但是您将函数定义为返回（QtreeNode），然后不返回（QtreeNode *）？

我在C ++中的方法有问题

3 个答案: