c++ - 访问基类的 protected 构造函数

Question

派生类可以在其ctor-initializer 中调用 protected 基类构造函数，但仅限于其自己的基类子对象，而不能在其他地方调用:

class Base {
  protected:
    Base() {}
};

class Derived : Base {
  Base b;
  public:
    Derived(): Base(),    // OK
               b() {      // error
        Base b2;          // error
    }
};

标准对此有何规定？这是 [class.protected]/1:

An additional access check beyond those described earlier in Clause 11 is applied when a non-static data member or non-static member function is a protected member of its naming class (11.2) As described earlier, access to a protected member is granted because the reference occurs in a friend or member of some class C. If the access is to form a pointer to member (5.3.1), the nested-name-specifier shall denote C or a class derived from C. All other accesses involve a (possibly implicit) object expression (5.2.5). In this case, the class of the object expression shall be C or a class derived from C. [ Example: ...

调用构造函数时是否涉及对象表达式？没有，是吗？那么标准中哪里描述了 protected 基类构造函数的访问控制？

Answer 1

protected 访问仅适用于您自己的当前对象类型 的父成员。您不能公开访问父类型的其他 对象的 protected 成员。在您的示例中，您只能访问默认基础构造函数作为 Derived 的一部分，而不是当它是作为 b 的独立对象时。

让我们分解一下您发布的标准 (11.4/1) 中的引述。我们假设标准中的 C 对应于您的 Derived 类:

An additional access check beyond those described earlier in Clause 11 is applied when a non-static data member or non-static member function is a protected member of its naming class (11.2).

所以基类构造函数在这里实际上是它的命名类 (B) 的一个非静态成员函数，所以这个条款到目前为止适用。

As described earlier, access to a protected member is granted because the reference occurs in a friend or member of some class C.

C 的成员(构造函数)所以我们在这里仍然很好。

If the access is to form a pointer to member (5.3.1), the nested-name-specifier shall denote C or a class derived from C.

这不是指向成员的指针，因此不适用。

All other accesses involve a (possibly implicit) object expression (5.2.5).

然后该标准断言所有其他可能的访问都必须涉及对象表达式。

In this case, the class of the object expression shall be C or a class derived from C.

最后，标准声明表达式的类必须是 C 或更进一步的派生类。在这种情况下，您的表达式 Base() 实际上是一个 C，调用父构造函数(将其视为 this->Base() . 表达式 b 显然是 Base 类型(这是成员 b 的显式声明类型，想想 this->b ->Base())。现在我们检查:Base 是 C 还是 C 的子级？不是, 所以这个代码是不合法的。