C++:为什么无法在派生类中访问 protected 构造函数？

Question

protected 成员应该可以从派生类访问。那么，为什么我在下面的代码中会出现编译错误？

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

class B : public A {
public:
    void g() { 
        A a; // <--- compiling error: "Protected function A::A() is not accessible ...". Why?
    }
};


int main() {
    B b;
    b.g();
}

我注意到有一个相关的帖子，但是那里的类是一个模板类。我的只是一个“常规”类(class)。

Why the derived class cannot access protected base class members?

Answer 1

protected members可以从派生类访问，但只能通过派生类访问。

A protected member of a class is only accessible

1. ...
2. to the members and friends (until C++17) of any derived class of that class, but only when the class of the object through which the protected member is accessed is that derived class or a derived class of that derived class:

因此，即使在派生类的成员函数中，也不能创建基类的独立对象。

换句话说，当前派生类实例的protected成员可以访问，而独立基类的protected成员不能访问。例如

class A {
protected:
    int x;
public:
    A() : x(0) {}
};

class B : public A {
public:
    void g() {
        this->x = 42; // fine. access protected member through derived class
        A a;
        a.x = 42;     // error. access protected member through base class
    }
};