我怎么能知道所有者对象的C ++中的地址?
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22-08-2019 - |
题
我想在C ++,我将在其他对象使用时,对象被销毁,以通知各个持有人通告类来创建。
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;
};
我的观点是因为我有一个致密的和复杂的对象图,我想避免存储拥有对象的地址的通知。有没有办法改变我的通知类,以便它可以从它自己的地址推断所拥有的对象的地址,并且会在编译时计算的?偏移
还要注意的是任何对象可以具有从同一类通知几个“所有者”,可能。
感谢。
解决方案
或者是这样的:
从您的通知继承和新增自有作为模板参数。然后你就可以有可用的通知内资方法:
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 )
{}
};
其他提示
看一看的 GoF的观测器的设计构成形式。
这将是一个讨厌破解,可能无法保证正常工作,但这里有一个想法的我不建议这样做
假设你有你的布局像你这样的描述:
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,它可以计算这样Owned
的地址(这是在Notifier
的构造执行):
Owned *p = reinterpret_cast<Owned *>(reinterpret_cast<char *>(this) - offsetof(Owned, _notifier));
基本上,假设是_notifier是在一些固定在所属类内的偏移。因此的拥有的地址等于_notifier
的地址减去相同的偏移。
再次,这是不确定的行为,我不建议,但能够工作。
FA“。的回答是一个良好的开端。但是,它不能解决具有相同类型的多个所有者的问题。一个解决方案是有通知店主,而不是单一的一个列表。这里是一个快速的实现,展现了主意:
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;
};
您可以使用这种方式:
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()) { }
};
在这里你有很多不同类型的业主的情况下,该解决方案可以变得相当难用。在这种情况下,你可能想看看升压元编程库(的 MPL ,融合一>),与你可能最终与让你做这样的东西一个代码:
class Owned : public Notifier<Owned, OwnerType1, OwnerType1, OwnerType2>
{
Owned(OwnerType1 & o1, OwnerType1 & o2, OwnerType2 & o3)
: base(o1,o2,o3)
};
然而,实施此解决方案将是比前一个时间稍长。
的解决方案的一部分。将已经从通知拥有继承。以这种方式,被破坏的对象的地址是简单的“本” ...
class Owned : public Notifier<Owner> {
public:
Owned(Owner* owner)
: Notifier<Owner>(owner)
{}
};
但如何从同一类处理多个“主人”?怎样才能从“同一类”?
继承几次由于 FA的回答,这里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!
我高度怀疑。没有办法的通知才知道它已经组成被使用。如果我做什么
class Foo
{
private:
Notifier _a, _b, _c;
}
我喜欢,虽然被证明是错误的,但我真的怀疑这是可行的,而不该通知明确给出更多的信息。