用派生类的类型覆盖函数参数类型 [英] Override function parameter type with type of derived class

问题描述

我计划使用一种采用自己类型的参数的方法来创建一个接口（而不是c ++中的虚拟基类）。

I plan to create an interface (rather a virtual base class in c++) with a method that takes an argument of the own type.

class Base {
public:
    virtual void seriousMethod(const Base &arg) = 0;
}

但是，派生类不应采用基类类型的参数，而应采用

The derived class should however take not an argument of the base class type but of the derived class type.

class Derived: public Base {
public:
    virtual void seriousMethod(const Derived &arg) { /* ... */ }
}

我知道吗？我需要对基类进行模板化（例如 Base< Derived> ）还是有更清洁的解决方案？

How would I realize this? Would I have to template the base class (e.g. Base<Derived>) or is there a cleaner solution?

推荐答案

您不能直接执行此操作。考虑这种情况：

You can't do this directly. Think about this case:

Base b;
Derived d;
Base& d_ref = d;
d_ref.seriousMethod(b);  // What happens here?

在编译时，变量 d_ref 具有静态类型 Base ，因此根据 Base 的定义，它应该能够使用 b 作为 seriousMethod 的参数。

At compile-time, the variable d_ref has static type Base, so according to the definition of Base, it should be able to take b as a parameter to seriousMethod.

但是在运行时，动态类型 d_ref 的值是 Derived ，因此根据 Derived ，它不能将 b 作为 seriousMethod 的参数。它不能将 b 转换为 Dervied ，因为它可能是直接的 Base 对象（如果 Base 不是抽象对象），或者它可能是从 Base 派生的其他一些类，与派生的不同。

But at runtime, the dynamic type of d_ref is Derived, so it according to the definition of Derived, it can't take b as a parameter to seriousMethod. It can't convert b to Dervied since it might be a straight Base object (if Base is not abstract), or it might be some other class derived from Base that is not the same as Derived.

您是正确的，假设唯一真正的解决方法是反复出现的模板模式，即将 Base 模板化并将 Dervied 定义为：

You are correct in assuming that the only real way to go about this is the curiously-recurring template pattern, i.e. templating Base and defining Dervied as:

class Derived : public Base<Derived> { ... }

这消除了上面说明的问题，因为每种类型均源自 Base< T> 将具有不同的基类，并且不会通过继承相互关联。

This removes the problem illustrated above, because each type derived from Base<T> will have a distinct base class, and will not be related to one another through inheritance.

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