c++ - 通过基类指针获取子类类型

Question

鉴于类布局(Base->Derived；Base->Derived2)，以及我将派生类的实例作为基类指针( Base* baseder2 = new Derived2 )，我希望能够实例化具有派生类型的 TemplClass 实例(类似于 sh = new TemplClass<Derived2>(baseder2) )。下面的代码实例化 sh = new TemplClass<Base>(baseder2) ，这会导致编译错误，因为 fn function未在类 Base 中声明。如何找出 baseder2 的派生类型指针，最好没有dynamic_cast？现实生活中的代码有很多 Base 后代，所以我想避免使用dynamic_cast 的 if 语句。我正在研究 boost::type_traits，但说实话，我不知道该怎么办。 模板函数template <typename T> BaseTemplClass* foo(T* t) 只是工厂对象的蹩脚借口。

最诚挚的问候， 多多尔

class Base
{
public:
  virtual ~Base(){}
};

class Derived : public Base
{
public:
  virtual ~Derived(){}
  void function()
  {
    std::cout<<"This is Derived"<<std::endl;
  }
};

class Derived2 : public Base
{
public:
  virtual ~Derived2(){}
   void function()
  {
    std::cout<<"This is Derived2"<<std::endl;
  }
};

class BaseTemplClass
{
public:
  virtual void Print() =0;
};

template <class Tmodel>
class TemplClass : public BaseTemplClass
{
public:
  TemplClass(Tmodel* m)
  {
    model = m;
  }
  void Print()
  {
    model->function();
    std::cout << " TemplClass"<<typeid(model).name() << std::endl;
  }
  Tmodel *model;
};

template <typename T> BaseTemplClass* foo(T* t)
{
  BaseTemplClass* sh;
  std::cout << "FOO: "<<typeid(t).name() << std::endl;
  sh = new TemplClass<T>(t);
  return sh;
}


int main(int argc, char **argv) 
{
    Derived* der = new Derived;
    Derived2* der2 = new Derived2;
    Base* baseder2 = new Derived2;

    BaseTemplClass* sh = foo(der);
    sh->Print();
    delete sh;

    sh = foo(der2);
    sh->Print();
    delete sh;


    sh = foo(baseder2);
    sh->Print();
    delete sh;

    delete der;
    delete der2;
    delete baseder2;
    return 0;
}

Answer 1

在Base中使function抽象更有意义，这样您就不需要知道它是哪个派生类，这样会更有意义吗？