从子子类或孙子访问基类

Question

我只想直截了当地说，因为当我学习继承时，我注意到我可以从已经派生的类继承，并且从最派生的类直接访问基类的成员。我认为以下内容表明了我的意思：

#include <iostream>

struct Base
{
    int member = 0;     // All members null
};

struct D1 : Base {};
struct D2 : Base {};
struct D3 : Base {};
struct D4 : Base {};

struct MostDerived : D1, D2, D3, D4 {};  // I know MostDerived now has 4 copies of Base


int main()
{
    MostDerived mostderived{};
    mostderived.D2::member = 2;                     // D2::Base::member = 2
    mostderived.D3::member = 3;                     // D3::Base::member = 3
    mostderived.D4::member = 4;                     // D4::Base::member = 4

    std::cout << mostderived.D1::member << '\n';        // Haven't touched D1, is still zero
    std::cout << mostderived.D1::Base::member << '\n';  // Equals 0, equivalent to above line
    std::cout << mostderived.Base::member << '\n';      // Read Base scope directly from most derived, equals zero

    mostderived.Base::member++;                         // Assign to member in Base scope directly from mostderived
                                                        // it now equals 1

    std::cout << mostderived.D1::member << '\n';        // But also does D1::member equal 1

    return 0;
}

所以通过mostderived.Base::member++ 我改变了mostderived.D1::member 的值。我只是想知道像这样访问Base 是否有意义，以及为什么它特别改变了D1 的成员副本。我描绘布局的方式是MostDerived 包含D1、D2、D3、D4，每个都包含Base，所以它看起来像下面这样。

Answer 1

所以通过执行 mostderived.Base::member++ 我改变了 mostderived.D1::member.

如果 'MostDerived' 继承自 D1-4，则

mostderived.Base::member++;
                  ^

模棱两可。

因为“MostDerived”具有这种层次结构

int Base::member
int Base::member
int Base::member
int Base::member

和Base 可以替换为 D1-4 中的任何一个。

---

如果你会使用虚拟继承

struct D1 : virtual Base {};
struct D2 : virtual Base {};
struct D3 : virtual Base {};
struct D4 : virtual Base {};

这将导致“MostDerived”只包含 一个 基类，所以它不会再模棱两可了

mostderived.Base::member++;

并且可以通过所有 D1-4 访问这个member

mostderived.D1::member = 1;
mostderived.D4::member = 4;

当然，因为所有 D 共享相同的变量，所以它对所有人都有相同的值，例如：

MostDerived mostderived{};
mostderived.Base::member = 1;
std::cout << mostderived.D1::member << mostderived.D2::member << mostderived.D3::member << mostderived.D4::member;

将打印1111。