在受保护的匿名联合中的匿名结构中定义的成员仍然公开可见

Question

我已经设法编译和运行 c++ 代码，尽管它不应该。

以下代码段不应编译：

template<typename T, size_t SIZE>
struct Vector {
    Vector(std::initializer_list<T> data) {
        std::copy(data.begin(), data.end(), this->data);
    }

    Vector(T(&data)[SIZE]) {
        std::copy(data, data + SIZE, this->data);
    }

protected:
#pragma pack(push, 1) //stores the alignment of aggregate types and sets it to 1 byte
    union {
        struct {
            T x, y, z, w;
        };
        T data[SIZE];
    };
#pragma pack(pop) //restore old data alignment
};

template<typename T>
struct Vector2 : public Vector<T, 2> {
    using Vector<T, 2>::Vector<T, 2>;

    Vector2(T x = 0, T y = 0) :
        Vector({ x, y }){}

    Vector2(const Vector& vec) :
        Vector(vec) {}

    using Vector::x;
    using Vector::y;
};

int main() {
    double floats[2]{ 2, 3 };
    Vector2<double> v{ floats };
    Vector<double, 2> c{ 5., 6. };

    std::cout << "v.x = " << v.x;
    //Is oke, v.x is visible here because of the public using statement

    std::cout << " c.x = " << c.x << "\n";
    //Is not oke, c is not a Vector2<double>. It is a Vector<double, 2> so its member x is protected and thus not visible from here.
}

输出： v.x = 2 c.x = 5

因此该程序不仅成功编译 link，而且运行并打印出合理的数据。我尝试将 c 的类型更改为 Vector<double, 3> 但这并没有改变任何东西。此外，成员 z 和 w 与 x 和 y 一样可见。但是，data 不可见（例如，std::cout << c.data[0]; 不会按预期编译）。

在这种情况下，Intellisense 比编译器更聪明，因为它成功检测到错误并提出错误。

我正在使用 Visual studio 2013.

PS:

附带问题：我在同一代码片段中发现了编译器的另一个怪癖。如果我更改以下行：

using Vector<T, 2>::Vector<T, 2>;

至：

using Vector<T, 2>::Vector;

我得到这个编译器错误：error C2886: 'Vector<T,0x02>' : symbol cannot be used in a member using-declaration

如果我把它改成：

using Vector::Vector;

编译器全部崩溃：fatal error C1001: An internal error has occurred in the compiler. see reference to class template instantiation 'Vector2<T>' being compiled。

这（例如它崩溃的事实）可能只是编译器中的一个错误，但如果有人知道，我仍然想知道为什么该行的两种替代形式都无法编译。

Answer 1

我必须进行一些更改才能让代码在 clang 上完全编译。看来visual c++非常放纵，允许非法的（或者说非标准的）语法。

这是更正后的程序：

#include <iostream>
#include <algorithm>

template<typename T, size_t SIZE>
struct Vector {
    Vector(std::initializer_list<T> data) {
        std::copy(data.begin(), data.end(), this->data);
    }

    Vector(T(&data)[SIZE]) {
        std::copy(data, data + SIZE, this->data);
    }

protected:
#pragma pack(push, 1) //stores the alignment of aggregate types and sets it to 1 byte
    union {
        struct {
            T x, y, z, w;
        };
        T data[SIZE];
    };
#pragma pack(pop) //restore old data alignment
};

template<typename T>
struct Vector2 : public Vector<T, 2> {
    using Vector<T, 2>::Vector;

    Vector2(T x = 0, T y = 0) :
    Vector<T, 2>({ x, y }){}

    Vector2(const Vector2& vec) :
    Vector<T, 2>(vec) {}

    using Vector<T, 2>::x;
    using Vector<T, 2>::y;
};

int main() {
    double floats[2]{ 2, 3 };
    Vector2<double> v{ floats };
    Vector<double, 2> c{ 5., 6. };

    std::cout << "v.x = " << v.x;
    //Is oke, v.x is visible here because of the public using statement

    std::cout << " c.x = " << c.x << "\n";
    //Is not oke, c is not a Vector2<double>. It is a Vector<double, 2> so its member x is protected and thus not visible from here.
}

这是修改后的（预期）错误：

./vec.cpp:66:33: error: 'x' is a protected member of 'Vector<double, 2>'
    std::cout << " c.x = " << c.x << "\n";
                                ^
./vec.cpp:37:15: note: declared protected here
            T x, y, z, w;
              ^
1 error generated.

在受保护的匿名联合中的匿名结构中定义的成员仍然公开可见

member defined in anonymous struct in protected anonymous union is still publicly visible

c++

c++11

visual-studio-2013

unions

anonymous-inner-class