要求对象产生于Heap之中
考虑如下代码:
classHeapClass { public: voidDestory() const {deletethis;} private: ~HeapClass(){} }; HeapClass* ptr = newHeapClass; ptr->Destory();
这样的调用真是很厉害,想生成非Heap对象都不成了。
对于继承和组合的情况不想多说了,比较无趣的说。
判断某个对象是否位于Heap内
考虑如下代码:
newHeapClass(* newHeapClass);
你觉得编译器应该怎么做?
1.调用operator new
2.调用Constructor
3.调用第二个operator new
4.调用第二个Constructor
但是可以让人足够惊讶,编译器对此并不做承诺,所以实际的实现可能是:
1.调用operator new
2.调用第二个operator new
3.调用Constructor
4.调用第二个Constructor
而VC6是这样实现的。
classHeapClass { private: void* operatornew[](size_tsize); typedefconstvoid * RawAddress; voidoperatordelete[](void* ptr); public: voidoperatordelete(void *ptr) { printf("delete "); ::operatordelete(ptr); m_address.erase(std::remove(m_address.begin(),m_address.end(),ptr),m_address.end()); return; } void* operatornew(size_tsize) { printf("new "); void * ptr = ::operatornew(size); m_address.push_back(ptr); returnptr; } HeapClass() { printf("Constructor! "); } HeapClass(constHeapClass&) { printf("copy Constructor! "); } virtualvoidDestory() const {deletethis;} virtual ~HeapClass() = 0; boolisOnHeap() const { // const void * rawAddress = dynamic_cast<const void *>(this); constvoid * rawAddress = (constvoid *)(this); std::deque<RawAddress>::iteratoriter = std::find(m_address.begin(),m_address.end(),rawAddress); returniter != m_address.end(); } private: staticstd::deque<RawAddress> m_address; }; HeapClass::~HeapClass(){} std::deque<HeapClass::RawAddress> HeapClass::m_address; classDHeapClass:publicHeapClass {};
我在VC6中写了这个Demo测试了一下,但是const void * rawAddress = dynamic_cast<const void *>(this);会出现异常,这让我觉得很郁闷,所以这个Demo只能支持普通的继承方式,不支持多种继承和虚拟继承。
禁止对象产生于heap之中
考虑如下代码:
classHeapClass { private: void* operatornew(size_tsize); void* operatornew[](size_tsize); voidoperatordelete(void *ptr); voidoperatordelete[](void* ptr); public: HeapClass(){printf("Constructor! ");} HeapClass(constHeapClass&){printf("copy Constructor! ");} public: ~HeapClass(){} };
这确实是比较简单的事情。 |