如何动态创建相关类型？

Question

How do I determine the type of a class that is related to another class dynamically?

I have figured out a solution, the only problem is that I ended up having to use a define that has to be used in all of the derived classes.

Is there a simpler way to do this that doesn't need upkeep for ever class that is added?

Things to note: both the class and the related class will always have their respective base class, the different classes can share a related class, and as in the example I would like the control class to own the view.

This is what I have now. The only thing I have to upkeep is the switch, but I would like to have it so I only need to add code insted of going back and changing it.

#include <iostream>
#include <string>

class model 
{
public:
  model( int id ) : id(id) {}

  int id;
};

class view 
{
public:
  view( model *m ) {}

  virtual std::string display() 
  {
     return "view";
  }
};

class otherView : public view 
{
public:
  otherView( model *m ) : view(m) {}

  std::string display() 
  {
     return "otherView";
  }
};


class control 
{
public:
  control( model *m ) : m_(m), v_( createRelated() ) {}

  ~control() 
  { 
     delete v_; 
  }
  std::string display() 
  {
     if ( v_ )
        return v_->display();
     return "No view";
  }

  view *createRelated() 
  {
     switch( m_->id )  
     {
     case 0:
       return new view( m_ );
     case 1:
       return new otherView( m_ );
     default:
       return NULL;
     }
  }

  model *m_;
  view  *v_;
};

int main( void ) {
  model m(0);
  model om(1);
  model nm(2);

  control c1( &m );
  control c2( &om );
  control c3( &nm );

  std::cout << c1.display() << std::endl;
  std::cout << c2.display() << std::endl;
  std::cout << c3.display() << std::endl;
}

Answer 1

A possible solution would be to change model to accept a pointer to a function that creates the related class, instead of passing an 'id':

typedef view* (*make_function)(model*);

class model
{
public:
    model(make_function a_make) : make_(a_make) {}
    view* make() const { return make_(this); }
...
private:
    make_function make_;
};

Each of view classes to provide a static method that creates an instance of itself:

class view
{
public:
    static view* make(model* m) {  return new view(m); }
};

class otherView: public view
{
public:
    static view* make(model* m) {  return new otherView(m); }
};

Then 'createRelated()' would become:

view* control::createRelated()
{
    return m_->make();
}

Example use:

model m1(&view::make);
model m2(&otherView::make);

Hope that helps.

Answer 2

您基本上是在寻找一种称为Virtual Constructor的技术。

Answer 3

您是否正在寻找工厂设计模式 ？

Answer 4

You might need some virtual functions. Even if you pass an object of type B derived from type A:

void f(A &objA) {
  objA.f();
}

int main() {
  B objB;
  f(objB);
  return 0;
}

if A::f() is defined to be virtual, then B::f() will be invoked. So you do not need to know that objA was an objB.