C ++：不完全类型

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我有以下代码：

template<class T> class RandomTreeNode { public: typedef typename RandomTreeFunction<T>::function_ptr function_ptr; RandomTreeNode(): left(NULL), right(NULL), threshold(0.0), is_a_leaf(false), data(NULL), function(0){ } void set_function(function_ptr function){this->function = function;} function_ptr get_function(){ return this->function;} void set_threshold(double threshold){this->threshold = threshold;} double get_threshold(){return threshold;} void create_left_child(){this->left = RandomTreeNode<T>();} //returning references so that they can be altered in a recursive tree build algo without excessive copying RandomTreeNode<T>& get_left_child(){return left;} void create_right_child(){this->right = RandomTreeNode<T>();} RandomTreeNode<T>& get_right_child(){return this->right;} bool is_leaf(){return this->is_a_leaf;} void mark_as_leaf(){this->is_a_leaf = true;} const std::vector<T> get_data(){ return data; } void set_data(std::vector<T>& data){ this->data = data; } private: RandomTreeNode<T> left; RandomTreeNode<T> right; double threshold; function_ptr function; std::vector<T> data; bool is_a_leaf; };

当编译时，我得到以下错误：RandomTreeNode< T& :: left'有不完整的类型。任何想法为什么？

推荐答案

因为它是你当前定义的类型。对于类型具有相同类型的成员（对于初学者，它将具有无限大小）没有意义。我想你想要的是具有指向 RandomTreeNode< T> 的指针，而不是直接实例。

Because it's the type you're currently defining. It doesn't make sense for a type to have a member of the same type (for starters, it would have an infinite size). I think what you want is to have pointers to RandomTreeNode<T>'s, not direct instances.