C++11 继承构造函数和访问修饰符

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.