Question

我正试图将现有的病媒阶层传入病媒。

class Vector
{
 public:
   float X,Y,Z;
};

设法使本类成员成为病媒,同时又不影响其他类别成员可变

class Vector
{
 public:
   union{
      float X,Y,Z;
      vector float vec4;
   };
};

但是,由于没有X、Y、Z号字眼,汇编错误。是否有其他办法实现变量?

For reference, the vector float type comes from the IBM™ Cell Broadband Engine™ Software Development Kit V3.0 for Multicore Acceleration.

Answer 1

struct VectorData {
  float X, Y, Z;
};

union VectorUnion {
  VectorData VecData;
  vector float vec4;
};


class Vector {
public:
  Vector() : X(u.VecData.X), Y(u.VecData.Y), Z(u.VecData.Z) {}
  float & X;
  float & Y;
  float & Z;
private:
  VectorUnion u;
};

这是我最接近的,我认为你能够获得标准C++—— 这使得Vector更大,要求你执行转让经营人的任务。

Answer 2

在标准C++中,没有办法使用工会。你只能读一下你以前写过的话。 http://code>X,Y,Z, 后读vec4 产生未界定的行为。

I would suggest to create a member function vector float toVector() const that will create the vector when needed. Or you may consider defining a member operator vector float() const.

Answer 3

你可以 do切地做你们想要的东西(利用你的法典中的辛子)。

One example correct way to do it:

union{
    float coords[4];
    vector float vec4;
};

And then you can pick off element by element.

Alternatively:

union{
    struct {
        float X, Y, Z;
    } v;
    vector float vec4;
};

但是,你通过<代码>v.X获取价值。

Answer 4

这应当做到。

#include <iostream>

using namespace std;

class Point
{
public:
    Point() { X = 0.0f; Y = 0.0f; Z = 0.0f; };
    ~Point() {};

    union
    {
        struct{
            float X; float Y; float Z;
        };
        float index[3];
    };

    float operator[](int it)
    {
        return index[it];
    };
};

void main()
{
    Point p;

    p.X = 1.0f;
    p.Y = 2.0f;
    p.Z = 3.0f;

    cout << p[0] << " " << p[1] << " " << p[2] << endl;

    p.index[0] = 4.0f;
    p.index[1] = 5.0f;
    p.index[2] = 6.0f;

    cout << p[0] << " " << p[1] << " " << p[2] << endl;

    cin.get();
}

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