虚拟函数可以被非虚函数覆盖吗？ [英] Can a virtual function be overriden by a non-virtual function?

问题描述

在此程式码中：

class Base {
public:
    virtual void method() = 0;
};

class Derived1 : public Base{
public:
    virtual void method() override { }
};

class Derived2 : public Base{
public:
    void method() override { }
};

Derived1 code> Derived2 ？

Is there any difference between Derived1 and Derived2?

推荐答案

> c ++ 11标准（草案n3337）第2点：

From section 10.3 Virtual functions of the c++11 standard (draft n3337) point 2:

如果虚拟成员函数vf在类Base在类Derived中直接或间接派生
来自Base，具有相同名称的成员函数vf，参数类型列表（8.3.5），cv-qualification和refqualifier
（或缺少声明为Base :: vf，然后Derived :: vf也是virtual（无论它是否是
这样声明的），它覆盖Base :: vf。

If a virtual member function vf is declared in a class Base and in a class Derived, derived directly or indirectly from Base, a member function vf with the same name, parameter-type-list (8.3.5), cv-qualification, and refqualifier (or absence of same) as Base::vf is declared, then Derived::vf is also virtual (whether or not it is so declared) and it overrides Base::vf.

因此 Derived2 :: method 也是 virtual ，即使它没有明确声明为这样。

So Derived2::method is also virtual, even though it is not explicitly declared as such.

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