具有多态模板参数的多态模板类 [英] Polymorphic templated classes with polymorphic template parameters

问题描述

在一个简单的设计中，类B多态地继承了类A. 模板类Base<T>具有一个T*成员，该成员用于进一步的操作. Derived<T>是Base<T>的多态继承. 允许这种对象创建的语法是什么?

In a simplistic design, a class B inherits a class A polymorphically. A templated class, Base<T> has a T* member that is used for further operations. A Derived<T> inherits from Base<T> polymorphically. What would be the syntax that allows this kind of object creation:

Base<A>* a = new Derived<B>();

为进一步参考，我使用的代码如下(当然，转换不会成功):

For further reference, the code I used looks like this (of course, the conversion does not succeed):

class A
{
public:
  A()
  {
    cout<< " A  ";
  }
  virtual void one()
  {
    cout<<" 1 ";
  }

};

class B: public A
{
public:
  B()
  {
    cout << " B  ";
  }
  void one()

  {
      cout << " 2 ";
  }
};

template <class T>
class Base
{

public:
    T* thing;
  Base()
  {
    cout<<"Base";
    thing = new T;
  }
  template <class S>
  Base(Base<S>* obj)
  {
    thing = obj->thing;
  }
  virtual void poly(){ thing->one();}
};
template <class T>
class Derived : public Base<T>
{
public:
  Derived()
  {
    cout << "DERIVED ";
  }
  virtual void poly(){ 
};
int main(int argc, char** argv)
{
  //Base<A>* a = (new Derived<B>());
  return 0;
}

为代码简洁起见，故意省略了虚拟析构函数和适当的内存管理.

Virtual destructors and proper memory management omitted on purpose for code brevity.

EDIT :此构造的唯一目的是将例如BaseTemplated<BasePolymorphic>*个指针的列表放在一起，而不是将BaseTemplated<Derived1>到BaseTemplated<DerivedN>用于a的所有N个子类.基本的多态类.

EDIT : the sole purpose of this construction would be to keep, say, a list of BaseTemplated<BasePolymorphic>* pointers together, and not use BaseTemplated<Derived1> to BaseTemplated<DerivedN> for all N subclasses of a base, polymorphic class.

推荐答案

首先，告诉我何时进入简单化"部分.

First, tell me when we get to the "simplistic" part.

剥离所有模板的一半内容，仅专注于您要进行多态处理的部分).最终，您将获得不同的类Base<A>和Base<B>. (后者是通过Derived<B>的推导得出的.)

Peel back all half of the templates to just concentrate on the part you're trying to polymorph). Eventually you have distinct classes Base<A> and Base<B>. (the latter through derivation of Derived<B>).

这两个继承都不来自其各自的模板参数.因此，A与B的关系(分层或其他方式)是不相关的. Base<A>和Base<B>是不同的且无关的，因此您尝试按原样进行的操作无效.实际上，即使 did 分别从A和B继承，您所希望的最好的情况就是A*指针，您在示例中没有使用该指针.

Neither of these inherit from their respective template parameters. Therefore the relationship (hierarchical or otherwise) of A to B is irrelevant. Base<A> and Base<B> are distinct and unrelated, and therefore what you're trying to do as-writen cannot work. In fact, even if the did inherit from a A and B respectively, the very best you could hope for is an A* pointer, which you aren't using in your sample.

如果没有另外显示，我会很高兴地删除它们，因为我真的很好奇.

And I'll gladly delete this if shown otherwise, because I'm genuinely curious.

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