将自身转换为派生类后可以访问c++基本私有方法吗？

Question

我在使用 CRTP（奇怪的重复模板模式）时发现了这一点。

template <typename T>
class Base {
private:
    void f() {
        //when T has its own f(), it calls that
        //when T doesn't have, it calls itself, which is invoked recursively
        //but i expect the compiler to complain that f() is private when T doesn't have its own f()
        static_cast<T*>(this)->f();
    }

public:
    void g() {
        f();
    }
};

class Derived : public Base<Derived> {};

我以为我理解public、protected 和private，但对于这种情况，我好像错了。任何解释表示赞赏！

Answer 1

这仅在阴影声明为public 时有效。见this example：

class Derived : public Base<Derived> {
private:
    void f() { }
};

void x(Derived* d) {
    d->g();
}

你得到：

<source>: In instantiation of 'void Base<T>::f() [with T = Derived]':
<source>:13:9:   required from 'void Base<T>::g() [with T = Derived]'
<source>:23:10:   required from here
<source>:8:9: error: 'void Derived::f()' is private within this context
         static_cast<T*>(this)->f();
         ^~~~~~~~~~~
<source>:19:10: note: declared private here
     void f() { }
          ^

如果函数没有在Derived 中被隐藏，则调用与this->Base<Derived>::f() 相同，这必须是合法的，因为Base 是唯一可以访问它的类。

您的困惑也可能来自访问看似不同的对象。请记住，访问修饰符按范围而不是实例限制访问。在Base 中声明的任何方法都可以任何Base 实例的私有成员，而不仅仅是this 的。