Question

我已经设法编译并运行 c ++ 代码，即使它不应该。

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，该程序不仅可以成功编译和链接，还可以运行并打印合理的数据。我尝试将c的类型更改为Vector<double, 3>，但这并未改变任何内容。此外，成员z和w也可见，就像x和y一样。但是，data不可见（例如std::cout << c.data[0];无法按预期编译）。

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抱怨它。

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>;

为：

using Vector<T, 2>::Vector;

我收到此编译错误：error C2886: 'Vector<T,0x02>' : symbol cannot be used in a member using-declaration

如果我将其更改为：

using Vector::Vector;

编译器与以下内容一起崩溃：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，但是，如果有人知道，我仍然想知道为什么这两行的两种替代形式都没有编译。

Answer 1

我必须做一些更改才能在clang上完全编译代码。似乎visual c ++是非常宽松的，允许非法（或者我应该说是非标准的）语法。

以下是更正后的计划：

#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.
}

以下是修改后的（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是公开可见的

1 个答案: