虚拟功能可以被非虚拟功能覆盖吗? [英] Can a virtual function be overridden by a non-virtual function?

问题描述

在此代码中:

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和Derived2之间是否有区别?

推荐答案

来自c ++ 11标准的 10.3虚拟函数部分(草稿n3337)第2点:

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

如果在Base类和Derived类中声明了虚拟成员函数vf，则直接或间接派生 从Base开始，成员函数vf具有相同的名称，parameter-type-list(8.3.5)，cv-qualification和refqualifier (或不存在)声明为Base :: vf，然后派生的:: vf也是虚拟的(无论是否 )，并且它会覆盖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.

所以Derived2::method也是virtual，即使未明确声明也是如此.

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

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