智能指针和dynamic_cast

如果这个问题已经在我查看时回答并且找不到答案，我提前道歉。

注意：这是一个家庭作业，所以如果你觉得回答不舒服，我完全理解。

我有以下几点：

PTR.H：

template<typename T>
class Ptr {
    T* address;
    size_t* counter;
    Ptr(T* address) : address(address), counter(new size_t(1)) { }
    Ptr(const Ptr& other) : address(other.address), counter(other.counter) 
    { 
        ++(*counter); 
    }
    virtual ~Ptr() { 
        if (0 == --(*counter)) { delete address; delete counter; } 
    }
    Ptr& operator=(const Ptr& right) {
        if (address != right.address) {
            if (0 == --(*counter)) { delete address; delete counter; }
            address = right.address;
            counter = right.counter;
            ++(*counter);
        }
        return *this;
    }
    T& operator*()  const { TRACE(address); return *address; }
    T* operator->() const { TRACE(address); return address;  }
    T* raw()        const { TRACE(addr); return addr;  }
};

main.cc：

#include <iostream>
#include "ptr.h"
using std::cout;
using std::endl;

class Base {
public:
    Base()  { std::cout << "Base()" << std::endl; }
    virtual ~Base() { std::cout << "~Base()" << std::endl; }
    virtual std::string str() { return "In Base::str()"; }
};
class Derived: public Base {
public:
    Derived()  { std::cout << "Derived()" << std::endl; }
    ~Derived() { std::cout << "~Derived()" << std::endl; }
    std::string str() { return "In Derived::str()"; }
};

int main() {
Ptr<Base> base(new Base());
Ptr<Derived> derived(new Derived());
Ptr<Base> pbase(0);
Ptr<Derived> pderived(0);
    // upcasting can be done like this, but is it the best way?
    pbase = *((Ptr<Base>*)(&derived));
    cout << pbase->str() << endl;   // outputs: "In Derived::str()"
    /* 
     * downcasting should be done using dynamic_casts
     * but how can I downcast here? 
     * what do I know so far:
     * 1. just because Derived is a subclass of Base does not mean Ptr<Derived> is a 
     * subclass of Ptr<Base> so there is no hierarchy between the two so I cannot 
     * really use dynamic_casts here
     * 2. The dynamic_cast I do use is sort of useless no? since pbase is a Ptr<Base>
     */
    pderived = *((Ptr<Derived>*)(dynamic_cast<Ptr<Base>*>(&pbase)));
    cout << pderived->str() << endl;

return 0;
}

现在，我们的目标是使用dynamic_cast来回切换，尽管我发现了许多关于智能指针的有趣花絮，但没有真正解释它是如何实现的。

我尝试只获取 pbase 的地址字段，然后将 pderived 初始化为具有该地址的新 Ptr，但当然我的引用计数都搞砸了。

我尝试创建一个新的 Ptr，其中包含对 pderived 计数器的引用，但后来我无法设置 pderived 的地址字段，所以我也卡在那里。

我告诉你们这些信息是因为：1.我想指出一点，在在线寻求帮助之前，我已经为此工作了很长时间，并且2.我想让你知道我已经尝试过什么。

我真的可以在这里使用一些建议。只是如何获得：

pderived = <SOMETHINGSOMETHING>pbase<SOMETHINGSOMETHING>

谢谢！