c++ - 类型转换问题

Question

假设您有基类 Unix_tree，以及派生类 Unix_tree_type1、Unix_tree_type2 等等。为什么我不能将 Unix_tree 转换为 Unix_tree_type...？有没有办法执行这样的类型转换？派生类包含抽象方法的实现，它们的目标是以不同的方式打印基类中包含的数据。谢谢。

class Unix_tree
    {
    public:
        Unix_tree () 
        {
            tree_it.tree = &tree;
        }

        Unix_tree(const Unix_tree& utree) : num_of_files(utree.num_of_files), tree(utree.tree), tree_it(utree.tree_it) 
        {
            tree_it.tree = &tree;
        };

        int num_of_files;
        map<int, pair<int, string> > names;
        vector<pair<int, int> > edges;

        Tree< pair<int, string> > tree;
        TreeIterator< pair<int, string> > tree_it;

        string input_format;
        string output_format;

        static Unix_tree* load_tree(istream& input_info);
        void load_names(istream& input_info);
        void load_tree_nodes(vector<int>& levels);
        Unix_tree* convert(string format);

        virtual void load(istream& input_info) = 0;
        virtual void print_function(const Tree<pair<int, string> >* node, const vector<bool>& visited ) = 0;
        void print_tree();
    };


    class Unix_tree_type1 : public Unix_tree
    {
    public:

        void load(istream& input_info);
        void print_function(const Tree<pair<int, string> >* node, const vector<bool>& visited);
    };

Answer 1

你可以施放指针，所以:

Unix_tree *p = ...;
Unix_tree_type1 *q = (Unix_tree_type1*)p;

但我认为你不想那样做。方法调用会自动执行您想要的操作，因此:

Unix_tree *p = ...; // points to an instance of Unix_tree_type1
p->load(...);       // calls Unix_tree_type1::load, not Unix_tree::load