c++ - 在派生类中使用指向基类的指针访问基类保护成员

Question

考虑以下代码:

#include <iostream>

using std::endl;
using std::cout;

template<typename T>
class B{
protected:
    T value;
    B* ptr;
public:
    B(T t):value(t), ptr(0){}
};

template<typename T>
class D: public B<T>{
public:
    void f();
    D(T t):B<T>(t){}
};

template<typename T>
void D<T>::f(){
    cout << this->value << endl;     //OK!
    this->ptr = this;
    cout << this->ptr->value << endl;      //error! cannot access protected member!!
    B<T>* a = this;
    cout << a->value <<endl;       //error! cannot access protected member!!
}


int main(){
    D<double> a(1.2);
    a.f();
    return 0;
}

看来基类的成员可以用this指针直接访问，其他指针不行。
编译器是否将它们视为不同的实例化？

Answer 1

是的，这是预期的行为。 Protected members基类可以在派生类中访问，但只能通过是派生类(或当前派生类的进一步派生类)的类型的对象(包括this)。这意味着您无法通过指向基类的指针访问 protected 成员。

A protected member of a class Base can only be accessed

1) ...

2) by the members of any class derived from Base, but only when operating on an object of a type that is derived from Base (including this)