具有混合继承修饰符的钻石继承（受保护/私有/公共）

发布于 2024-12-28 12:32:56 字数 960 浏览 0 评论 0原文

假设我们有类 A,B,C,D，其中 A 是基础，B,C 是介于两者之间，D 是在钻石模型中派生的。

注意：

B 类 在私有模式下虚拟继承A 类，

C 类继承保护模式下的 virtualy A 类。

class A
{
public:
    int member;  // note this member
};
class B :
    virtual private A // note private 
{

};
class C :
    virtual protected A // note protected
{

};
class D :
    public B, // doesn't metter public or whatever here
    public C
{

};

int main()
{
    D test;
    test.member = 0; // WHAT IS member? protected or private member?
    cin.ignore();
    return 0;
}

现在，当我们创建class D 的实例时，成员会是什么？私人的还是受保护的哈哈？

图 2：

如果我们这样做会怎样：

class B :
    virtual public A // note public this time!
{

};
class C :
    virtual protected A // same as before
{

};

我想 member 在第二个示例中将是公开的，不是吗？

原文

let's say we have class A,B,C,D where A is base, B,C are between and D is derived in diamond model.

NOTE:

class B inherits virtualy class A in private mode,

class C inherita virtualy class A in protected mode.

class A
{
public:
    int member;  // note this member
};
class B :
    virtual private A // note private 
{

};
class C :
    virtual protected A // note protected
{

};
class D :
    public B, // doesn't metter public or whatever here
    public C
{

};

int main()
{
    D test;
    test.member = 0; // WHAT IS member? protected or private member?
    cin.ignore();
    return 0;
}

now when we make an instance of class D what will member be then? private or protected lol?

Figure No2:

what if we make it so:

class B :
    virtual public A // note public this time!
{

};
class C :
    virtual protected A // same as before
{

};

I suppose member will be public in this second example isn it?

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半衾梦 2025-01-04 12:32:56

§11.6 多路访问 [class.paths]

如果可以通过多重继承图通过多个路径访问某个名称，则访问权限是提供最多访问权限的路径的访问权限。 [示例：
class W { public: void f(); };
A 类：私有虚拟 W { }；
B 类：公共虚拟 W { }；
C 类：公共 A，公共 B {
   无效 f() { W::f(); } // 好的
};
由于 W::f() 可沿公共路径通过 B 供 C::f() 使用，因此允许访问。 —结束示例]

我认为我不需要添加任何其他内容，但另请参阅此缺陷报告（已关闭为“不是缺陷”）。

§11.6 Multiple access [class.paths]

If a name can be reached by several paths through a multiple inheritance graph, the access is that of the path that gives most access. [ Example:
class W { public: void f(); };
class A : private virtual W { };
class B : public virtual W { };
class C : public A, public B {
   void f() { W::f(); } // OK
};
Since W::f() is available to C::f() along the public path through B, access is allowed. —end example ]

I think I don't need to add anything else, but see also this defect report (which was closed as "not a defect").

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