c++ - 通过基类指针获取子类类型

给定类的布局(Base-> Derived; Base-> Derived2)，以及我将实例作为基类指针(Base* baseder2 = new Derived2)保留到派生类的事实，我希望能够使用实例化TemplClass实例。派生类型(类似于sh = new TemplClass<Derived2>(baseder2))。下面的代码实例化
sh = new TemplClass<Base>(baseder2)，由于未在类function中声明fn Base，导致编译错误。如何找出baseder2指针的派生类型，最好是没有dynamic_cast？现实生活中的代码有许多Base后代，因此我想避免使用带有dynamic_cast的if语句。我一直在研究boost::type_traits，但老实说，不要做什么。
模板函数template <typename T> BaseTemplClass* foo(T* t)只是工厂obj的la脚借口。

最好的祝福，
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;
}

最佳答案

在您的情况下真的需要模板类吗？在给定的示例中，您可以将方法“function”添加到基类，并将其虚拟化。

但是，如果仍然需要模板类，则可以执行以下操作:

class Base
{
public:
  virtual ~Base(){}
  virtual BaseTemplClass* createTemplClass() = 0;
};

class Derived : public Base
{
public:
  virtual ~Derived(){}
  void function()
  {
    std::cout<<"This is Derived"<<std::endl;
  }
  virtual BaseTemplClass* createTemplClass()
  {
    return new TemplClass<Derived>( this );
  }
};

class Derived2 : public Base
{
public:
  virtual ~Derived2(){}
  void function()
  {
    std::cout<<"This is Derived2"<<std::endl;
  }
  virtual BaseTemplClass* createTemplClass()
  {
    return new TemplClass<Derived2>( this );
  }
};


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