关于c ++：C ++ 11继承构造函数和访问修饰符

C++11 inheriting constructors and access modifiers

假设如下布局：

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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;
};

我可以在这里指定Derived的构造函数为公共的吗？VC++给出以下错误：

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'

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也就是说，它忽略了继承构造函数上方的访问修饰符。

这是功能的限制吗？Base类有一个公共构造函数是没有意义的，因为它永远不能直接实例化(由于是纯虚拟方法)。

根据12.9/4，"继承建设者"，在说using X::X时，

A constructor so declared has the same access as the corresponding constructor in X.

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所以继承的构造函数也是protected。