3

我有几个孩子班都有相同的父母。每个子类都可以使用包含在父对象中的一些数据来构造。我想使用基础对象中包含的信息将一个孩子变成另一个孩子（不修改基础对象）。

目前它的实现如下例所示：

#include <iostream>

using namespace std;

class Data {};
class base
{
public:
    base()  {}
    base(Data input) : data(input)  {}
    virtual ~base() {   cout << "Deleting :" << this->name() << endl;   }
    template<class T> static base* CastToDerrived(base* object)
    {
        T* output = new T(object->data);
        delete object;
        return output;
    }
    virtual const char* name()  {return "base";}
    Data data;
};

class derrived1 : public base
{
public:
    derrived1() {}

    derrived1(Data input): base(input){}
    ~derrived1(){cout << "Deleting :" << this->name() << endl;}
    const char* name(){return "derrived1";}
};

class derrived2 : public base
{
public:
    derrived2(){}
    derrived2(Data input): base(input){}
    ~derrived2(){cout << "Deleting :\t" << this->name() << endl;}
    const char* name(){return "derrived2";}
};

int main(int argc, char *argv[])
{
    base* object = new derrived1();
    cout << "Created :\t"<<object->name()<<endl;
    object = base::CastToDerrived<derrived2>(object);
    cout << "Casted to :\t"<<object->name()<<endl;
}

哪个输出：

Created :   derrived1
Deleting :derrived1
Deleting :base
Casted to : derrived2

但是，这需要销毁并再次创建基类，我想避免这种情况 - 我想销毁 derrived1，使用 base 构造 derrived2，但保持基类完整。最好的方法是什么？

（有几个派生类，基类为所有提供了通用接口，一些派生类以后会在不修改基类的情况下添加）。

4

1 回答 1

0

您不能销毁派生类并保留基类，因为它是同一个实例。我们可以使用以下代码演示这一点

#include <iostream>

using namespace std;

class Data {};
class base
{
public:
    base()
    {
        printf("base pointer = %08X\n", this);
    }
};

class derrived1 : public base
{
public:
    derrived1()
    {
        printf("derrived1 pointer = %08X\n", this);
    }
};

int main(int argc, char *argv[])
{
    base* object = new derrived1();
    delete object;
    return 0;
}

输出将显示 base an derived1 是同一个指针，另一方面，您使用一些棘手的代码来实现您想要的。这个想法是使用指向 Data 类的指针而不是变量。您的代码将如下所示

#include <iostream>

using namespace std;

class Data {};
class base
{
public:
    base(){data = new Data}
    base(Data* input) : data(input)  {}
    virtual ~base() {   cout << "Deleting :" << this->name() << endl;   }
    template<class T> static base* CastToDerrived(base* object)
    {
        T* output = new T(object->data);
        delete object; 
        return output;
    }
    virtual const char* name()  {return "base";}
    Data* data;
};

class derrived1 : public base
{
public:
    derrived1() {}

    derrived1(Data* input): base(input){}
    ~derrived1(){cout << "Deleting :" << this->name() << endl;}
    const char* name(){return "derrived1";}
};

class derrived2 : public base
{
public:
    derrived2(){}
    derrived2(Data* input): base(input){}
    ~derrived2(){cout << "Deleting :\t" << this->name() << endl;}
    const char* name(){return "derrived2";}
};

int main(int argc, char *argv[])
{
    base* object = new derrived1();
    cout << "Created :\t"<<object->name()<<endl;
    object = base::CastToDerrived<derrived2>(object);
    cout << "Casted to :\t"<<object->name()<<endl;
}

我希望这会有所帮助。

于 2013-04-13T20:05:59.040 回答