如何从指向多态基类的指针复制/创建派生类实例

问题描述

我一直在为这种问题苦苦挣扎，所以我决定在这里提问.

I have been struggling with this kind of problem for a long time, so I decided to ask here.

class Base {
  virtual ~Base();
};
class Derived1 : public Base { ... };
class Derived2 : public Base { ... };
...

// Copies the instance of derived class pointed by the *base pointer
Base* CreateCopy(Base* base);

该方法应该返回一个动态创建的副本，或者至少将对象存储在某些数据结构中的堆栈上，以避免返回临时地址"问题.

The method should return a dynamically created copy, or at least store the object on stack in some data structure to avoid "returning address of a temporary" problem.

实现上述方法的天真的方法是在一系列 if 语句中使用多个 typeid 或 dynamic_cast 来检查每个可能的派生类型，然后使用 new 运算符.还有其他更好的方法吗?

The naive approach to implement the above method would be using multiple typeids or dynamic_casts in a series of if-statements to check for each possible derived type and then use the new operator. Is there any other, better approach?

P.S.:我知道，使用智能指针可以避免这个问题，但我对没有一堆库的简约方法感兴趣.

P.S.: I know, that the this problem can be avoided using smart pointers, but I am interested in the minimalistic approach, without a bunch of libraries.

推荐答案

您在基类中添加一个 virtual Base* clone() const = 0; 并在您的派生类中适当地实现它.如果你的 Base 不是抽象的，你当然可以调用它的复制构造函数，但这有点危险:如果你忘记在派生类中实现它，你会得到(可能不需要的)切片.

You add a virtual Base* clone() const = 0; in your base class and implement it appropriately in your Derived classes. If your Base is not abstract, you can of course call its copy-constructor, but that's a bit dangerous: If you forget to implement it in a derived class, you'll get (probably unwanted) slicing.

如果您不想复制该代码，您可以使用 CRTP 惯用语通过模板实现功能:

If you don't want to duplicate that code, you can use the CRTP idiom to implement the function via a template:

template <class Derived>
class DerivationHelper : public Base
{
public:
  virtual Base* clone() const
  {
    return new Derived(static_cast<const Derived&>(*this)); // call the copy ctor.
  }
};

class Derived1 : public DerivationHelper <Derived1> { ... };
class Derived2 : public DerivationHelper <Derived2> { ... };

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