c++ - 为什么不能将继承的 protected 构造函数公开？

Question

考虑:

class A
{
protected:
    A(int) {}
    void f(int) {}

public:
    A() {}
};

class B : public A
{
public:
    using A::A;
    using A::f;
};

int main()
{
    B().f(1); // ok
    B(1); // error: 'A::A(int)' is protected within this context
}

为什么不能将继承的 protected构造函数设为 public，而不能将继承的 protected成员函数设为？

Answer 1

与其他成员不同，引入inherited constructor的using声明的可访问性将被忽略。

[namespace.udecl]/19，

(强调我的)

A synonym created by a using-declaration has the usual accessibility for a member-declaration. A using-declarator that names a constructor does not create a synonym; instead, the additional constructors are accessible if they would be accessible when used to construct an object of the corresponding base class, and the accessibility of the using-declaration is ignored.