私有使用基构造函数的声明不是私有的

Question

基础构造函数的 using 声明是私有的，但仍然可以构造 class。为什么？

对于 operator[] 的 using 声明，辅助功能的工作方式不同，必须是 public.

#include <vector>

template<typename T>
class Vec : std::vector<T>
{
private:
    using std::vector<T>::vector;       // Works, even if private. Why?
public:
    using std::vector<T>::operator[];   // must be public
};

int main(){
    Vec<int> vec = {2, 2};
    auto test = vec[1];
}

如果我希望构造函数是私有的怎么办？可以用 using 声明来完成吗？

Answer 1

基础 class 构造函数的使用声明保持与基础 class 相同的可访问性，无论基础 class 的可访问性如何。来自 [namespace.udecl]:

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

^强调

用简单的英语来说，from cppreference：

It has the same access as the corresponding base constructor.

如果你想"inherited"构造函数是私有的，你必须手动指定构造函数。您不能使用 using 声明来执行此操作。

Answer 2

using reference 指出继承的构造函数

has the same access as the corresponding base constructor.

它进一步暗示了这背后的基本原理：

It is constexpr if the user-defined constructor would have satisfied constexpr constructor requirements. It is deleted if the corresponding base constructor is deleted or if a defaulted default constructor would be deleted

显然，您不能显式地 constexpr 或 delete 继承的构造函数，因此这些特性只是被继承。访问级别也是如此。

私有使用基构造函数的声明不是私有的

Private using declaration of base constructor is not private

c++

inheritance

constructor

using-declaration