C++ 在c+中使用友元类的私有构造函数处理类构造函数+；

在单词搜索示例程序中，我有抽象基类查询、句柄类查询和派生类WordQuery

句柄类是抽象基类和派生类的朋友，它的构造函数应该使用派生类的私有构造函数

以下是这些类的代码：

class Query_base

{
    friend class Query;
protected:
    typedef TextQuery::line_no line_no;
    virtual ~Query_base() { };
private:
    // eval returns the |set| of lines that this Query matches
    virtual std::set<line_no> eval (const TextQuery&) const = 0;
    // display prints the query
    virtual std::ostream& display (std::ostream& = std::cout) const = 0;
};

// handle class to manage the Query_base inheritance hierarchy
class Query
{
    // these operators need access to the Query_base* constructor
    friend Query operator~(const Query &);
    friend Query operator|(const Query&, const Query&);
    friend Query operator&(const Query&, const Query&);
public:
    Query(const std::string&); // build a new WordQuery
    // copy control to manage pointers and use counting
    Query(const Query &c): q(c.q), use(c.use) { ++*use; }
    ~Query() { decr_use(); }
    Query operator=(const Query&);

    // interface functions: will call corresponding Query_base operations
    std::set<TextQuery::line_no> eval (const TextQuery &t) const 
      { return q->eval(t); }
    std::ostream &display(std::ostream&os) const { return q->display(os); }
private:
    Query(Query_base *query): q(query), use (new std::size_t(1)) { }
    Query_base *q;
    std::size_t *use;
    void decr_use()
    {
        if (--*use == 0)
        {
            delete q;
            delete use;
        }
    }
};

class WordQuery: public Query_base
{
    friend class Query; // Query uses the WordQuery constructor
    WordQuery(const std::string &s): query_word(s) { }
    // concrete class: WordQuery defines all inherited pure virtual function
    std::set<line_no> eval(const TextQuery &t) const
        { return t.run_query(query_word); }
    std::ostream& display (std::ostream &os) const
        { return os << query_word; }
    std::string query_word; // word for which to search
};

但编译器不让我这么做

请让我知道我做错了什么

对于那些可能面临类似问题的人，构造函数应该这样编写：

Query::Query(const std::string &s): q(WordQuery(s)) { };

Query::Query(const std::string &s)
{ q = new WordQuery(s); use = new std::size_t(1); }

---

q

是一个指针，您需要

new

WordQuery@TartanLlama谢谢。