Question

让我们说我想在下面的课后得到所有东西，我不假设任何进一步的新的重载或删除：

class Object
{
public:
        static map<std::string, unsigned int> typeDeltaMap;
        void* operator new(size_t size)
        {
                void* p = ::new char[size];
                const std::string type = typeid(this).name(); //compile error
                cout << "new type=" << type << endl;
                ++typeDeltaMap[type];
        }
        void operator delete(void* p)
        {
                ::delete(p);
                const std::string type = typeid(this).name(); //compile error
                cout << "delete type=" << type << endl;
                --typeDeltaMap[type];
        }
};

我想最终得到像这样的东西

class A : public Object
{
        public:
                virtual ~A(){}
};

class B : public Object
{
        public:
                virtual ~B(){}
};

int main()
{
        A* a = new A();//prints new type=A
        B* b = new B();//prints new type=B
        delete a;//prints delete type=A
        delete b;//prints delete type=B
}

我没有工作，因为新的和删除是静态的，但有一些特殊的静态，因为它们同时是虚拟的。我的问题是，是否有某种方法可以获得这些信息？

Answer 1

正如@Someprogrammerdude建议我使用CRTP来满足我的需求

template<typename T>
class Object
{
public:
        ;
        void* operator new(size_t size)
        {
                const std::string type = typeid(T).name();
                cout << "new type=" << type << endl;
                void* p = ::new char[size];
                return p;
        }
        void operator delete(void* p)
        {
                const std::string type = typeid(T).name();
                cout << "delete type=" << type << endl;
                ::delete(p);
        }
};

class A : public Object<A>
{
public:
virtual ~A(){}
};

class B : public Object<B>
{
public:
virtual ~B(){}
};

int main()
{
        A* a = new A();
        B* b = new B();
        delete a;
        delete b;
}

从operator new / delete中获取有关对象的RTTI信息

1 个答案: