Question

我正在使用以下使用std :: shared_ptr（C ++ 11）的类结构：

#include <string>
#include <iostream>
#include <vector>
#include <memory>

//class Member is given by a 3rd party library, CAN'T modify its structure!
class Member {//no instance of this class allowed
   public:
     Member() {};//should never be called
     virtual ~Member() = 0;//pure virtual distructor;
     virtual void foo() {
       std::cout<<"Member"<<std::endl;
     }
};
Member::~Member() {} //need to define destructor because of child classes 

class ChildMember : public Member {
  public:
     ChildMember() {};
     virtual void foo() {
         std::cout<<"Child Member"<<std::endl;
     }

     virtual void foo2() {
        std::cout<<"unique foo in child"<<std::endl;
     }
};

class Base {
   public:
     Base() {};

   virtual
   std::shared_ptr< Member >
   get_var() {return var;}

   virtual void 
   set_var ( std::shared_ptr< Member > v) { var = v;}

   std::shared_ptr< Member > var;
};


class ChildBase : public Base {
  public:
     ChildBase() {
        //var = std::make_shared<ChildMember>();
     };

  virtual
  std::shared_ptr< ChildMember >
  get_var() {return var;} //(2) : try to comment

  virtual void 
  set_var ( std::shared_ptr< ChildMember > v) { var = v;}

  std::shared_ptr< ChildMember > var;
};

void func(std::shared_ptr<Base> b) {
    b->get_var()->foo();// process input using polymorphism
}

int main() 
{
    std::shared_ptr<ChildBase> cb( std::make_shared<ChildBase>());
    cb->set_var (std::make_shared<ChildMember>() );
    cb->get_var()->foo2();//(3) want to use unique functions of ChildBase;
    cb->var->foo2();      //can access directly as well;
    //cb->var = std::make_shared<ChildMember>();
    func(cb);
}

我尝试做的是设计两个类（Base和ChildBase），其中每个类都有自己的成员Member / ChildMember。完全使用ChildBase对象及其ChildMember类型的成员对象后，我将其传递给接收func的函数Base，并且应该使用多态性调用成员foo的正确var。

问题：的（1） 在上面的变体中，编译器抱怨：

 overriding 'virtual std::shared_ptr<Member> Base::get_var()'
     get_var() {return var;}`

（2） 如果我评论ChildBase get_var的实现，则此函数将被视为Base，从而返回指向Member但没有foo2的指针。

（3） 我还可以评论cb->get_var()->foo2()。整个事情都会编译，但似乎没有调用ChildMember::foo()，因为没有输出Child Member;

有些东西（希望）有些混乱，但我看不到什么。你能帮忙解决一下吗？

EDIT1 ：

根据Abhijit Kadam的回答，以下代码完全编译：

#include <string>
#include <iostream>
#include <vector>
#include <memory>

class Member {//no instance of this class allowed
   public:
     Member() {};
     virtual ~Member() = 0;//pure virtual distructor;
     virtual void foo() {
       std::cout<<"Member"<<std::endl;
     }
};
Member::~Member() {} //need to define destructor for child classes 

class ChildMember : public Member {
  public:
     ChildMember() {};
     virtual void foo() {
        std::cout<<"Child Member"<<std::endl;
     }

     void foo2() {
        std::cout<<"unique foo in child"<<std::endl;
     }
};

class Base {
   public:
     Base() {};

     std::shared_ptr< Member >
     get_var() {return var;}

     void set_var ( std::shared_ptr< Member > v) { var = v;}

     std::shared_ptr< Member > var;
};


class ChildBase : public Base {
  public:
     ChildBase() {
       //var = std::make_shared<ChildMember>();
     };  

     std::shared_ptr< ChildMember > var;
};

void func(std::shared_ptr<Base> b) {
    b->get_var()->foo();// process input using polymorphism
}

 void func_vec( std::vector< 
              std::shared_ptr<Base>
           > vec) {
     for (unsigned int i=0;i<vec.size();i++)
          vec[i]->get_var()->foo();
  }

int main() 
{
   std::shared_ptr<ChildBase> cb( std::make_shared<ChildBase>());
   cb->set_var (std::make_shared<ChildMember>() );

   func(cb);

   std::vector< std::shared_ptr<Base>> vec;
   vec.push_back(cb);
   func_vec(vec);

   cb->var->foo2();

   /*std::shared_ptr<ChildMember> ptr(std::dynamic_pointer_cast<ChildMember>(cb->get_var()) );
   if (ptr) {
      ptr->foo2();
   } */
}

EDIT2 在下面添加了答案。

Answer 1

virtual
std::shared_ptr< Member >
get_var() {return var;}

virtual
std::shared_ptr< ChildMember >
get_var() {return var;} //(2) : try to comment

这是错误的重写。 n3337 10.3 / 7

覆盖函数的返回类型应与被覆盖的返回类型相同与函数类的函数或协变。如果函数D :: f覆盖函数B :: f，则如果函数的返回类型满足以下条件，则它们是协变的：

- 两者都是指向类的指针，都是对类的左值引用，或者两者都是对它们的右值引用类

- B :: f的返回类型中的类与返回类型D :: f中的类相同，或者是返回类型为D :: f
的类的明确且可访问的直接或间接基类
- 指针或引用在D :: f的返回类型中具有相同的cv-qualification和类类型与B :: f的返回类型中的类类型具有相同的cv资格或更少的cv资格。

您的情况下所有条件都失败了。您可以使用原始指针或引用，例如http://liveworkspace.org/code/4eQWBI $ 1 但我认为你应该重写你的界面而不是使用foo2，如果它不是虚拟的。

Answer 2

要解决您的问题，请参阅此代码的修改和注释版本：

#include <string>
#include <iostream>
#include <vector>
#include <memory>

class Member {//no instance of this class allowed
public:
    Member() {};
    virtual ~Member() = 0;//pure virtual distructor;
    virtual void foo() {
        std::cout<<"Member"<<std::endl;
    }
};
Member::~Member() {} //need to define destructor for child classes 

class ChildMember : public Member {
public:
    ChildMember() {};
    virtual void foo() {
        std::cout<<"Child Member"<<std::endl;
    }

    virtual void foo2() {
        std::cout<<"unique foo in child"<<std::endl;
    }
};

class Base {
public:
    Base() {};

    // maybe you should have a virtual dtor here
    virtual ~Base();

    // note: NOT virtual
    std::shared_ptr< Member >
    get_var() {return var;}

    // also NOT virtual
    void set_var ( std::shared_ptr< Member > v) { var = v;}

    std::shared_ptr< Member > var;
};


class ChildBase : public Base {
public:
    ChildBase() {
        //var = std::make_shared<ChildMember>();
    };

    // non-virtual version which casts var to the correct type
    std::shared_ptr< ChildMember >
    get_var() {return std::dynamic_pointer_cast<ChildMember>(var);}

    // note: no var here, use the one from Base
};

void func(std::shared_ptr<Base> b) {
    b->get_var()->foo();// process input using polymorphism
}

int main() 
{
    std::shared_ptr<ChildBase> cb( std::make_shared<ChildBase>());
    cb->set_var (std::make_shared<ChildMember>() );
    cb->get_var()->foo2();//(3) want to use unique functions of ChildBase;

    // This can't work:
    //cb->var->foo2();      //can access directly as well;

    //cb->var = std::make_shared<ChildMember>();
    func(cb);
}

Answer 3

由于覆盖函数的返回类型不能有所不同，我将使用动态强制转换来实现解决方案。因此，不需要为`get_var和set_var'提供虚函数。除非该类将由其他类进一步派生，否则也不需要foo2为虚拟。

int main() 
{
    std::shared_ptr<ChildBase> cb( std::make_shared<ChildBase>());
    cb->set_var (std::make_shared<ChildMember>() );

    shared_ptr<ChildMember> ptr(dynamic_pointer_cast<ChildMember>(cb->get_var()) );
    if (ptr) {
        ptr->foo2();
    } 
}

class Member {//no instance of this class allowed
   public:
     Member() {};
     virtual ~Member() = 0;//pure virtual distructor;
     virtual void foo() {
       std::cout<<"Member"<<std::endl;
     }
};
Member::~Member() {} //need to define destructor for child classes 

class ChildMember : public Member {
  public:
     ChildMember() {};
     virtual void foo() {
         std::cout<<"Child Member"<<std::endl;
     }

     void foo2() {
        std::cout<<"unique foo in child"<<std::endl;
     }
};

class Base {
   public:
     Base() {};

   std::shared_ptr< Member >
   get_var() {return var;}

   void set_var ( std::shared_ptr< Member > v) { var = v;}

   std::shared_ptr< Member > var;
};


class ChildBase : public Base {
  public:
     ChildBase() {
        //var = std::make_shared<ChildMember>();
     };  

  std::shared_ptr< ChildMember > var;
};

void func(std::shared_ptr<Base> b) {
    b->get_var()->foo();// process input using polymorphism
}

Answer 4

shared_ptr 不是指针，因此协方差不适用。

通常情况下，我对协方差的使用持怀疑态度。一世通常会发现派生类更清晰，不会改变返回类型，并在需要时提供新功能返回指向派生函数的指针。在你的情况下，我会说这是必不可少的，因为你实际上有两个指针（最终可能指向不同的对象）。

Answer 5

过了一会儿，我虽然以下代码可能是一个很好的解决方案：

#include <string>
#include <iostream>
#include <vector>
#include <memory>

class Member {//no instance of this class allowed
   public:
     Member() {};
     virtual ~Member() = 0;//pure virtual distructor;
     virtual void foo() {
       std::cout<<"Member"<<std::endl;
     }
};
Member::~Member() {} //need to define destructor for child classes 

class ChildMember1 : public Member {
  public:
     ChildMember1() {};
     virtual void foo() {
        std::cout<<"Child Member 1"<<std::endl;
     }

     void foo1() {
        std::cout<<"unique foo in child 1"<<std::endl;
     }
};

class ChildMember2 : public Member {
  public:
     ChildMember2() {};
     virtual void foo() {
        std::cout<<"Child Member 2"<<std::endl;
     }
};

class ChildChildMember2 : public ChildMember2 {
   public:
      ChildChildMember2() {};
      virtual void foo() {
         std::cout<<"Child-Child Member 2"<<std::endl;
      }
};

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

     virtual
     std::shared_ptr< Member >
     get_var() = 0; //purely abstract class

};

class ChildBase1 : public Base {
  public:
     ChildBase1() {
       var = std::make_shared<ChildMember1>();
     };

     virtual
     std::shared_ptr< Member >
     get_var() {return var;}

     std::shared_ptr< ChildMember1 > var;
};

class ChildChildBase : public ChildBase1 {
   public:
      ChildChildBase() {
         var2 = std::make_shared< ChildChildMember2>();
      }

   virtual
   std::shared_ptr< Member >
   get_var() { return var2; }

   std::shared_ptr< ChildChildMember2 > var2;
};


class ChildBase2 : public Base {
  public:
     ChildBase2() {
       var = std::make_shared<ChildMember2>();
     };  

     virtual
     std::shared_ptr< Member >
     get_var() {return var;}

     std::shared_ptr< ChildMember2 > var;
};

void func(std::shared_ptr<Base> b) {
    b->get_var()->foo();// process input using polymorphism
}

int main() 
{
   std::shared_ptr<ChildBase1> cb1( std::make_shared<ChildBase1>());
   cb1->var->foo1();
   func(cb1);

   std::shared_ptr<ChildBase2> cb2( std::make_shared<ChildBase2>());
   func(cb2);

   std::shared_ptr<ChildChildBase> ccb( std::make_shared<ChildChildBase>());
   func(ccb);

}

我认为它实现了我没有并行继承或动态转换的情况。它还有超过1 lvl的继承。

输出结果为：

unique foo in child 1
Child Member 1
Child Member 2
Child-Child Member 2

类的继承和多态及其成员（shared_ptr，C ++ 11）

5 个答案: