Question

假设你有这个：

class foo {
public:
  virtual int myFunc() = 0;
  ///...
  virtual bool who()   = 0; // don't want to implement this
};

class bar : public foo {
public:
  int myFunc() {return 3;}
  //...
  bool who() {return true;} // don't want to implement this
};

class clam : public foo {
public:
  int myFunc() {return 4;}
  //...
  bool who() {return false;} // don't want to implement this
};


int main() {

  std::vector<foo*> vec (2, NULL);
  vec[0] = new bar();
  vec[1] = new clam();

  // copy vec and allocate new ptrs as copies of the data pointed to by vec[i]
  std::vector<foo*> vec2 (vec.size(), NULL);
  for ( int i=0; i<vec.size(); ++i ) {

    // obviously not valid expression, but it would be nice if it were this easy
    //vec2[i] = new foo(*vec[i]); 

    // the hard way of copying... is there easier way?
    if (vec[i]->who()) {
      vec2[i] = new bar ( * static_cast<bar* >(vec[i]) ) ;
    } else {
      vec2[i] = new clam( * static_cast<clam*>(vec[i]) );
    }

  }

  return 0;
}

我想要的是让编译器在其簿记中查找并根据存储的* vec [i]类型分配/复制vec2 [i]的简单方法。解决方法是创建一个虚函数，它基本上返回一个值，指定* vec [i]的类型，然后根据它进行条件分配。

Answer 1

一种常见的方法是这样的：

class foo {
public:
  virtual foo* clone() = 0;
};

class bar : public foo {
public:
  virtual bar* clone() { return new bar(*this); }
};

class clam : public foo {
public:
  virtual clam* clone() { return new clam(*this); }
};

Answer 2

一种方法是使用动态强制转换来确定对象的类型，例如在此完成（Finding the type of an object in C++）。但最简单的方法可能是使用typeid。

（假设您想要保持使用类型作为限定词的方式，否则我会推荐Joachim或Igor作为更好的选择:)）

Answer 3

您可以使用dynamic_cast进行向下转换并测试类型

bar* pbar = dynamic_cast<bar*>(vec[i])
if (pbar) {
  vec2[i] = new bar ( * static_cast<bar* >(vec[i]) ) ;
} else {
  vec2[i] = new clam( * static_cast<clam*>(vec[i]) );
}

请参阅dynamic_cast中的更多信息 http://www.cplusplus.com/doc/tutorial/typecasting/

C ++从抽象基类指针复制数据？

3 个答案: