派生类的虚拟赋值运算符未被调用

时间:2011-10-30 18:50:29

标签: c++ assignment-operator

我对C ++很陌生,我正试图掌握虚拟任务。下面的程序包含一个带有两个数据成员的抽象基类,以及一个带有一个数据成员的派生类。当我设置一个派生对象的抽象指针时,程序使用operator =的抽象版本而不是派生版本,即使它们都被声明为“虚拟”。我在这里做错了什么?

提前致谢,

#include <iostream>
#include <cstring>

class Abstract
{
  protected:
        char * label;
        int rating;
    public:
        Abstract(const char * l = "null", int r = 0);
        virtual Abstract & operator=(const Abstract & rs);
        virtual ~Abstract() { delete [] label; }
        virtual void view() const = 0;

};

class Derived : public Abstract
{
    private:
        char * style;
    public:
        Derived(const char * s = "none", const char * l = "null",
                  int r = 0);
        ~Derived() { delete [] style; }
        virtual Derived & operator=(const Derived & rs);
        virtual void view() const;

};

Abstract::Abstract(const char * l , int r )
{
    label = new char[std::strlen(l) + 1];
    std::strcpy(label, l);
    rating = r;
}

Abstract & Abstract::operator=(const Abstract & rs)
{
    if (this == &rs)
            return *this;
    delete [] label;
    label = new char[std::strlen(rs.label) + 1];
    std::strcpy(label, rs.label);
    rating = rs.rating;
    return *this;
}

Derived::Derived(const char * s, const char * l, int r)
         : Abstract(l, r)
{
    style = new char[std::strlen(s) + 1];
    std::strcpy(style, s);
}

Derived & Derived::operator=(const Derived & hs)
{
    if (this == &hs)
        return *this;
    Abstract::operator=(hs);
    style = new char[std::strlen(hs.style) + 1];
    std::strcpy(style, hs.style);
    return *this;
}

void Derived::view() const
{
    std::cout << "label: " << label << "\nrating: "
              << rating << "\nstyle: " << style;
}

int main ()
{
    using namespace std;
    char label[20], style[20];
    int rating;


    cout << "label? ";
    cin >> label;
    cout << "rating? ";
    cin >> rating;
    cout <<"style? ";
    cin >> style;

    Derived a;
    Abstract * ptr = &a;
    Derived b(style, label, rating);
    *ptr = b;
    ptr->view();

    return 0;
}

2 个答案:

答案 0 :(得分:0)

C ++不允许您使用协变参数类型覆盖虚函数。您的派生运算符根本不会覆盖抽象赋值运算符,它定义了一个完全正交的运算符,只是因为它具有相同的运算符名称。

你必须小心创建这样的函数,因为如果两个实际的派生类型不一致,那么几乎可以肯定,赋值将是荒谬的。我会重新考虑是否可以通过替代方法更好地满足您的设计需求。

答案 1 :(得分:0)

这有点旧,但万一其他人偶然发现它:

要添加Mark的答案,您可以通过实施

来实现
Derived & operator=(const Abstract & rs);

在这种情况下,您可能需要通过投射使用rsdynamic_cast<const Derived &>(rs)
当然这应该只是小心翼翼。 完整的实施将是:

Derived & Derived::operator=(const Abstract & hs)
{
    if (this == &hs)
        return *this;
    Abstract::operator=(hs);
    style = new char[std::strlen(dynamic_cast<const Derived &>(hs).style) + 1];
    std::strcpy(style, dynamic_cast<const Derived &>(hs).style);
    return *this;
}