我可以将指针传递给超类，但创建子类的副本吗? [英] Can I pass a pointer to a superclass, but create a copy of the child?

问题描述

我有一个函数，它接受一个指向超类的指针并对其执行操作.但是，在某些时候，该函数必须对输入的对象进行深层复制.有什么办法可以进行这样的复制吗?

I have a function that takes a pointer to a superclass and performs operations on it. However, at some point, the function must make a deep copy of the inputted object. Is there any way I can perform such a copy?

我想到将函数设为模板函数并简单地让用户传递类型，但我希望 C++ 提供更优雅的解决方案.

It occurred to me to make the function a template function and simply have the user pass the type, but I hold out hope that C++ offers a more elegant solution.

推荐答案

SpaceCowboy 提出了惯用的 clone 方法，但忽略了 3 个关键细节:

SpaceCowboy proposes the idiomatic clone method, but overlooked 3 crucial details:

class Super
{
public:
  virtual Super* clone() const { return new Super(*this); }
};

class Child: public Super
{
public:
  virtual Child* clone() const { return new Child(*this); }
};

1. clone 是一个 const 方法
2. clone 返回指向当前类的指针，而不是基类
3. clone 返回当前对象的副本

1. clone is a const method
2. clone returns a pointer to the current class, not the base class
3. clone returns a copy of the current object

第二个非常重要，因为它可以让用户受益于这样一个事实，即有时您拥有比 Super* 更多的类型信息.

The 2nd is very important, because it allows use to benefit from the fact that sometimes you have more type information than just a Super*.

另外，我通常更喜欢 clone 提供一个副本，而不仅仅是一个相同类型的新对象.否则，您正在使用 Exemplar 模式来构建新对象，但您没有正确克隆并且名称具有误导性.

Also, I usually prefer clone to provide a copy, and not merely a new object of the same type. Otherwise you're using an Exemplar pattern to build new objects, but you're not cloning proper and the name is misleading.

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