抽象类和派生类的输出运算符(<<)

Question

我有2个类,一个抽象基类和派生类.但由于某种原因,我不能正确地重载两个输出运算符.

这是基类:

class team
{
    char* team_name;
    int games_played;

public:
    team(const char* tname);
    virtual ~team();
    virtual void update_lpoints(const int win_loss)=0;
    friend std:: ostream& operator<< (std :: ostream& out, team& T);    
};

std:: ostream& operator<< (std :: ostream& out, team& T);

这是输出运算符:

std:: ostream& operator<< (std :: ostream& out, team& T)
{
    return (out<<T.team_name<<" "<<T.games_played);
}

派生类:

class BasketTeam : public team
{
    int league_points;
    int points_for;
    int points_against;

public:
        BasketTeam(const char* tname);  
        ~BasketTeam();

        friend std:: ostream& operator<< (std :: ostream& out, BasketTeam& T);  
};

std:: ostream& operator<< (std :: ostream& out, BasketTeam& T); 

这是派生类的输出运算符:

std:: ostream& operator<< (std :: ostream& out, BasketTeam& T)
{
    out<<T.get_name()<<"    "<<T.get_played_games()<<"  "<<T.league_points<<"   "<<T.points_for<<"  "<<T.points_against<<endl;
    return out;
}

当我创建对象并尝试打印它时,我只能让基类看起来不是派生类.

team* tt = new BasketTeam ("ran");
cout<<*tt;

提前致谢.

Answer 1

根据operator <<参数的静态类型,在编译时选择重载.既然tt是静态类型team,那operator <<就是它所使用的.

如果希望对象的动态类型确定输出,则必须使用其他一些技术.例如,您可以team包含一个虚print函数,并覆盖它BasketTeam.然后,让一个operator <<拿一个team& t,然后打电话t.print().这将print根据t您要查找的动态类型调用该方法.