问题描述
假设如下布局:
class Base
{
protected:
Base(P1 p1, P2 p2, P3 p3);
public:
virtual void SomeMethod() = 0;
}
class Derived : public Base
{
public:
using Base::Base;
public:
virtual void SomeMethod() override;
};
我可以在这里将 Derived 的构造函数指定为 public 吗?VC++ 给出以下错误:
Should I be able to specify Derived's constructor as public here? VC++ gives the following error:
无法访问在类派生"中声明的受保护成员
编译器在此处生成了Derived::Derived"[指向使用 Base::Base行]
见派生"的声明
cannot access protected member declared in class 'Derived'
compiler has generated 'Derived::Derived' here [points to the using Base::Base line]
see declaration of 'Derived'
即它忽略了继承构造函数上方的访问修饰符.
i.e. it's ignoring the access modifier above the inherited constructor.
这是功能的限制吗?Base 类具有公共构造函数没有任何意义，因为它永远无法直接实例化(由于纯虚方法).
Is this a limitation of the feature? It doesn't make any sense for the Base class to have a public constructor, as it can never be instantiated directly (due to the pure virtual method).
推荐答案
根据12.9/4，继承构造函数"，当说using X::X时，
According to 12.9/4, "Inheriting constructors", when saying using X::X,
如此声明的构造函数与 X 中相应的构造函数具有相同的访问权限.
A constructor so declared has the same access as the corresponding constructor in X.
所以继承的构造函数也是protected.
So the inherited constructor is also protected.
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