父函数和子函数的重载 - 如何访问父函数

Question

这是我想做的：

class Msg {
    int target;
public:
    Msg(int target): target(target) { }
    virtual ~Msg () { }
    virtual MsgType GetType()=0;
};

inline std::ostream& operator <<(std::ostream& ss,Msg const& in) {
    return ss << "Target " << in.target;
}

class Greeting : public Msg {
    std::string text;
public:
    Greeting(int target,std::string const& text) : Msg(target),text(text);
    MsgType GetType() { return TypeGreeting; }
};

inline std::ostream& operator <<(std::ostream& ss,Greeting const& in) {
    return ss << (Msg)in << " Text " << in.text;
}

不幸的是，这不起作用，因为倒数第二行的 Msg 转换失败，因为 Msg 是抽象的。然而，我希望代码能够仅在一处输出父级的信息。这样做的正确方法是什么？谢谢！

编辑：抱歉，为了清楚起见，这是这一行 return ss << （消息）在<< 《正文》<< in.text;我不知道怎么写。

Answer 1

尝试ss<<(Msg const&)in。
也许你必须让操作员成为 Greeting 类的友元。

#include "iostream"
#include "string"

typedef enum {  TypeGreeting} MsgType;

class Msg {
    friend inline std::ostream& operator <<(std::ostream& ss,Msg const& in);

    int target;
public:
    Msg(int target): target(target) { }
    virtual ~Msg () { };
        virtual MsgType GetType()=0;
};

inline std::ostream& operator <<(std::ostream& ss,Msg const& in) {
    return ss << "Target " << in.target;
}

class Greeting : public Msg {
    friend inline std::ostream& operator <<(std::ostream& ss,Greeting const& in);

    std::string text;
public:
    Greeting(int target,std::string const& text) : Msg(target),text(text) {};
    MsgType GetType() { return TypeGreeting; }

};

inline std::ostream& operator <<(std::ostream& ss,Greeting const& in) {
    return ss << (Msg const&)in << " Text " << in.text;
}

int _tmain(int argc, _TCHAR* argv[])
{
    Greeting grt(1,"HELLZ");
    std::cout << grt << std::endl;
    return 0;
}

不是很好的设计，但解决了你的问题。