什么是C ++中clone（）的最好的签名？

问题描述

正如Scott Myers所写，您可以利用C ++的类型系统中的放松来声明clone（），以返回指向所声明的实际类型的指针：

As Scott Myers wrote, you can take advantage of a relaxation in C++'s type-system to declare clone() to return a pointer to the actual type being declared:

class Base
{
    virtual Base* clone() const = 0;
};

class Derived : public Base
{
    virtual Derived* clone() const
};

编译器检测到clone（）返回指向对象类型的指针，

The compiler detects that clone() returns an pointer to the type of the object, and allows Derived to override it to return a pointer to derived.

希望clone（）返回一个智能指针，意味着转移所有权语义，如下所示：

It would desirable to have clone() return a smart pointer that implies transfer of ownership semantics, like the following:

class Base
{
   virtual std::auto_ptr<Base> clone() const = 0;
};

class Derived : public Base
{
    virtual std::auto_ptr<Derived> clone() const;
};

不幸的是，约定的松弛不适用于模板化的智能指针，编译器不允许

Unfortunately, the relaxation of the conventions does not apply to templated smart pointers, and the compiler will not allow the override.

所以，它似乎有两个选项：

So, it seems I am left with two options:

2. 有clone（）返回一个智能基础指针，并且客户端使用dynamic_cast来处理它。

这些方法中的其中一种是首选吗？

Is one of these approaches preferred? Or is there a way for me to eat my transfer of ownership semantics and have my strong type safety too?

推荐答案

这取决于你的所有权转让方式吗？您的用例。如果你认为你需要在你知道的动态类型的一个派生对象上调用 clone （记住， clone 是允许复制而没有知道动态类型），那么你应该可能返回一个哑指针并加载到智能指针在调用代码。如果没有，那么你只需要返回一个smart_ptr，所以你可以随时返回它在所有覆盖。

It depends on your use case. If you ever think you will need to call clone on a derived object whose dynamic type you know (remember, the whole point of clone is to allow copying without knowing the dynamic type), then you should probably return a dumb pointer and load that into a smart pointer in the calling code. If not, then you only need to return a smart_ptr and so you can feel free to return it in all overrides.

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