Question

我知道,在你想要宣布多变功能时,你必须宣布基类功能虚拟。

class Base
{
public:
    virtual void f();
};

我的问题是,你是否要求宣布继承的班级功能为虚拟功能,即使它预计儿童的行为与“淡化”一样?

class Child : public Base
{
public:
    void f();
};

Answer 1

无,你不需要重新宣布这一功能。

a) 基准类别中的“虚拟”栏目:

struct A
{
   void foo();          //not virtual
};
struct B : A
{
   virtual void foo();  //virtual
}
struct C : B
{
   void foo();          //virtual
}

Answer 2

宣布(f)为“儿童虚拟”有助于阅读儿童的定义。它是有用的文件。

Answer 3

一旦基类上下限,即标明为virtual。所有其他压倒性做法都是如此。虽然您不必将这一职能标记为virtual。为了文件目的,我倾向于这样做。

As of the last part: even if it s expected that Child behaves as if it were "sealed"?, if you want to seal the class, you can actually do it in C++11 (this was not fully implementable in C++03 generically) by creating a seal class like so:

template <typename T>
class seal {
   seal() {}
   friend T;
};

之后继承您的密封等级(CRTP):

class Child : public Base, virtual seal<Child> {
// ...
};

The trick is that because of the use of virtual inheritance, the most derived type in the hierarchy must call the virtual base constructor (int this case seal<Child>), but that constructor is private in the templated class, and only available to Child through the friend declaration.

在C++中,你要么必须建立<条码>。您希望密封的每类物品的类型,或采用不提供perfect密封的通用方法(可以更改)

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