如何知道C++中的所有者对象的地址?

看一看 GoF Observer Design Patter .

这将是一场灾难 肮脏的 黑客,可能不能保证工作,但这里有一个想法 .

假设您有如下描述的布局:

template <class Owner>
class Notifier<Owner> {
public:
  Notifier(Owner* owner);
  ~Notifier(); // Notifies the owner that an object is destroyed
};

class Owner;

class Owned {
public:
  Owned(Owner* owner);
private:
  Notifier<Owner> _notifier;
};

如果 _notifier 知道它的名字,它可以计算 Owned 的地址(在 Notifier

Owned *p = reinterpret_cast<Owned *>(reinterpret_cast<char *>(this) - offsetof(Owned, _notifier));

基本上,假设_通知程序位于所属类中的某个固定偏移量。因此,拥有的地址等于 _通知者 的地址减去相同的偏移量。

再一次,这是我不推荐的未定义行为,但可能有效。

fa.'s answer 这是一个好的开始。但是,它不能解决具有相同类型的多个所有者的问题。一种解决方案是让通知程序存储所有者列表,而不是单个所有者列表。下面是一个快速实现,以展示这一想法:

template <typename Owner, typename Owned>
class Notifier
{
  protected:
    Notifier()
    {}

    // Constructor taking a single owner
    Notifier(Owner & o) 
    { 
        owners.push_back(&o); 
    }

    // Constructor taking a range of owners
    template <typename InputIterator>
    Notifier(InputIterator firstOwner, InputIterator lastOwner)
        : owners(firstOwner, lastOwner) {}

    ~Notifier()
    {
        OwnerList::const_iterator it = owners.begin();
        OwnerList::const_iterator end = owners.end();
        for ( ; it != end ; ++it)
        {
            (*it)->notify(static_cast<Owned*>(this));
        }
    }

    // Method for adding a new owner
    void addOwner(Owner & o) 
    { 
        owners.push_back(&o); 
    }

private:
    typedef std::vector<Owner *> OwnerList;
    OwnerList owners;
};

class Owner;

class Owned : public Notifier<Owner, Owned>
{
    typedef Notifier<Owner, Owned> base;

    //Some possible constructors:
    Owned(Owner & o) : base(o) { }

    Owned(Owner & o1, Owner & o2)
    {
        base::addOwner(o1); //qualified call of base::addOwner
        base::addOwner(o2); //in case there are other bases
    }

    Owned(std::list<Owner*> lo) : base(lo.begin(), lo.end()) { }
};

如果您有许多不同类型的所有者,则此解决方案可能会变得非常难以使用。在本例中,您可能需要查看boost元编程库( MPL , Fusion

class Owned : public Notifier<Owned, OwnerType1, OwnerType1, OwnerType2>
{
    Owned(OwnerType1 & o1, OwnerType1 & o2, OwnerType2 & o3) 
        : base(o1,o2,o3)
};

但是,实现此解决方案将比上一个解决方案稍微长一点。

或者像这样:

从通知程序继承并添加Owned as template参数。然后,您可以在通知程序中使用自己的方法:

template < class Owner , class Owned >
class Notifier
{
public:
    Notifier(Owner* owner)
    {}

    Owned * owned()
    { return static_cast< Owned * >( this ); }

    ~Notifier()
    {
        // notify owner with owned()
    }
};

class Owner
{};

class Owned : public Notifier< Owner , Owned >
{
public:
    Owned( Owner * owner ) : Notifier< Owner , Owned >( owner )
    {}
};

部分解决方案是从通知程序继承。这样,被销毁对象的地址就是“This”。。。

class Owned : public Notifier<Owner> {
public:
  Owned(Owner* owner) 
    : Notifier<Owner>(owner)
  {}
};

但是如何处理来自同一类的多个“所有者”?如何从“同一类”继承多次?

fa's answer ,以下是我一直在寻找的解决方案:

#include <iostream>

template <class Owner, class Owned, int = 0>
class Notifier {
public:
  Notifier(Owner* owner)
    : _owner(owner)
  {}
  ~Notifier() {
    _owner->remove(owned());
  }
  Owned * owned(){ 
    return static_cast< Owned * >( this ); 
  }

private:
  Owner* _owner;
};

class Owner {
public:
  void remove(void* any) {
    std::cout << any << std::endl;
  }
};

class Owned : public Notifier<Owner,Owned,1>, Notifier<Owner,Owned,2> {
public:
  Owned(Owner* owner1, Owner* owner2)
    : Notifier<Owner,Owned,1>(owner1)
    , Notifier<Owner,Owned,2>(owner2)
  {}
};

int main() {
  std::cout << sizeof(Owned) << std::endl;
  Owner owner1;
  Owner owner2;
  Owned owned(&owner1, &owner2);
  std::cout << "Owned:" << (void*)&owned << std::endl << std::endl;
}

谢谢

我对此深表怀疑。通知程序无法知道它已在合成中使用。如果我这样做呢

class Foo
{
private:
  Notifier _a, _b, _c;
}