Question

我偶然发现了这堂课：

class Vec3f
{
    ...
    float x, y, z;
    ...
};

inline float operator[](const int index) const
{
    return (&x)[index];
}

inline float& operator[](const int index)
{
     return (&x)[index];
}

该类使用[]来访问x，y，z值，就像在数组中一样 v [0]是x中的值，v [1]是y中的值，v [2]是z中的值，但是

return语句如何工作？
读取它是否正确：“从x的地址开始获取索引指定的地址中的值”？
Do（＆amp; x）必须在括号中，否则它会返回x [index]的地址值，不是吗？

Answer 1

从技术上讲，这不是有效的代码。

但是发生了什么：

// Declare four variables
// That are presumably placed in memory one after the other.
float x, y, z;

在代码中：

return (&x)[index];

// Here we take the address of x (thus we have a pointer to float).
// The operator [] when applied to fundamental types is equivalent to 
// *(pointer + index)

// So the above code is
return *(&x + index);
// This takes the address of x. Moves index floating point numbers further
// into the address space (which is illegal).
// Then returns a `lvalue referring to the object at that location`
// If this aligns with x/y/z (it is possible but not guaranteed by the standard)
// we have an `lvalue` referring to one of these objects.

很容易使这项工作变得合法：

class Vec3f
{
    float data[3];
    float& x;
    float& y;
    float& z;

    public:
        float& operator[](const int index) {return data[index];}

        Vec3f()
            : x(data[0])
            , y(data[1])
            , z(data[2])
        {}
        Vec3f(Vec3f const& copy)
            : x(data[0])
            , y(data[1])
            , z(data[2])
        {
            x = copy.x;
            y = copy.y;
            z = copy.z;
        }
        Vec3f& operator=(Vec3f const& rhs)
        {
            x = rhs.x;
            y = rhs.y;
            z = rhs.z;
            return *this;
        }
};

在这种情况下，这个operator []重载如何工作？

1 个答案: