将自身转换为派生 class 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）时发现了这个。

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，看来我错了。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 中隐藏，调用与 this->Base<Derived>::f() 相同，必须是合法的，因为 Base 是唯一的 class可以访问它。

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

将自身转换为派生 class 后可以访问 C++ 基本私有方法吗？

c++ base private method can be accessed after casting itself to derived class?

c++

private

crtp