C++ 类访问级别

Question

假设我有两个类(class)。一个叫做点:

class Point{
    public:
        Point(): x_(0), y_(0) {}
    protected: 
        int x_, y_;
    };

然后我有另一个类，它派生自 Point:

class Point3D : public Point {
public:
    Point3D(): Point(), z_(0){}
    double distance(Point3D pt, Point base) const;
protected:
    int z_;
};

double Point3D::distance(Point3D pt, Point base) const
{
    int dist_x, dist_y;
    dist_x = base.x_ - x_;
    dist_y = base.y_ - y_;

    return sqrt(dist_x * dist_x +
                dist_y * dist_y);
}

然后我得到类似这样的错误:base.x_ is protected within this context。但是 Point3D 对 Point 的访问级别是 public，加上 Point 中的 x_ 数据成员是 protected 。所以它应该没有错误，对吧？有人可以帮我解决这个问题吗？

Answer 1

But the access level of Point3D to Point is public

这并不完全正确。只有通过派生类指针/引用访问基类的 protected 成员时，派生类才能访问它们。

double Point3D::distance(Point3D pt, Point base) const
{
    int dist_x, dist_y;
    dist_x = base.x_ - x_; // x_ is same as this->x_. Hence it is OK.
                           // base is not a Point3D. Hence base.x_ is not OK.
    dist_y = base.y_ - y_;

    return sqrt(dist_x * dist_x +
                dist_y * dist_y);
}

---

允许通过基类指针/引用访问基类的protected 成员将允许您以导致意外后果的方式更改对象。

假设你有:

class Base
{
   protected:
     int data;
};

class Derived1 : public Base
{
    void foo(Base& b)
    {
       b.data = 10; // Not OK. If this were OK ... what would happen?
    }
};

class Derived2 : public Base
{
};

int main()
{
    Derived1 d1;
    Derived2 d2;
    d1.foo(d2); 
}

如果

       b.data = 10;

在 Derived1::foo 中被允许，您可以修改 d2 的状态，Derived2 的实例 派生 1。这种后门修改不是理想的行为。