OpenCV的cvMat的内存结构是什么?

问题描述

想象一下我有以下东西:

Imagine I have the following:

CvMat* mat = cvCreateMat(3,3,CV_16SC3)

这是第3通道的3x3整数矩阵.

This is a 3x3 matrix of integers of channel 3.

现在，如果您查看OpenCV文档，则会发现以下内容是cvMat的减速情况:

Now if you look at OpenCV documentation you will find the following as the deceleration for cvMat:

typedef struct CvMat {

int type;
int step;

int* refcount;

union
{
    uchar* ptr;
    short* s;
    int* i;
    float* fl;
    double* db;
} data;

union
{
    int rows;
    int height;
};

union
{
    int cols;
    int width;
};
} CvMat;

现在，我想玩一下data.ptr，它是指向存储在cvMat中的数据的指针.但是，我很难理解内存的布局方式.如果我有一个3通道矩阵，这是如何工作的?对于一个频道，它很简单，因为它只是一个简单的MxN矩阵，其中M为行，N为cols.但是对于3个频道，是否有3个这些MxN矩阵的??有人可以告诉我如何通过data.ptr初始化3通道矩阵以及如何访问这些值吗?谢谢.

Now, I want to play around with the data.ptr, which is the pointer to the data stored in cvMat. However, I'm having a hard time understanding how the memory is layed out. If I have a 3 channel matrix, how does this work? For one channel its simple because it's justa simple matrix of MxN where M is rows and N is cols. However for 3 channel, are there 3 ofthese MxN matrix's?? Can someone show me how I would go about initalizing a 3 channel matrix via data.ptr and how to access these values please? Thank you.

推荐答案

此网页是对OpenCV 1.1的出色介绍.我建议使用最新版本的 Open CV 2.0 ，该版本具有处理图像，矩阵等的常规Mat类.与OpenCV 1.1不同.

This webpage is an excellent introduction to OpenCV 1.1. I would recommend using the latest version, Open CV 2.0 which has a general Mat class which handles images, matrices, etc. unlike OpenCV 1.1.

以上网页详细介绍了以下用于在多通道图像中访问元素的方法:

The above webpage has detailed the following methods for element access in multi-channel images:

间接访问:(一般，但效率低下，可以访问任何类型的图像)

对于多通道浮动(或字节)图像:

For a multi-channel float (or byte) image:

IplImage* img=cvCreateImage(cvSize(640,480),IPL_DEPTH_32F,3);
CvScalar s;
s=cvGet2D(img,i,j); // get the (i,j) pixel value
printf("B=%f, G=%f, R=%f\n",s.val[0],s.val[1],s.val[2]);
s.val[0]=111;
s.val[1]=111;
s.val[2]=111;
cvSet2D(img,i,j,s); // set the (i,j) pixel value

直接访问:(有效访问，但容易出错)

对于多通道浮动图片:

IplImage* img=cvCreateImage(cvSize(640,480),IPL_DEPTH_32F,3);
((float *)(img->imageData + i*img->widthStep))[j*img->nChannels + 0]=111; // B
((float *)(img->imageData + i*img->widthStep))[j*img->nChannels + 1]=112; // G
((float *)(img->imageData + i*img->widthStep))[j*img->nChannels + 2]=113; // R

使用指针进行直接访问:(在有限的假设下简化且有效的访问)

对于多通道浮动图像(假定4字节对齐):

For a multi-channel float image (assuming a 4-byte alignment):

IplImage* img  = cvCreateImage(cvSize(640,480),IPL_DEPTH_32F,3);
int height     = img->height;
int width      = img->width;
int step       = img->widthStep/sizeof(float);
int channels   = img->nChannels;
float * data    = (float *)img->imageData;
data[i*step+j*channels+k] = 111;

使用c ++包装器直接访问:(简单而有效的访问)

为单通道字节图像，多通道字节图像和多通道浮动图像定义c ++包装器:

Define a c++ wrapper for single-channel byte images, multi-channel byte images, and multi-channel float images:

template<class T> class Image
  {
    private:
    IplImage* imgp;
    public:
    Image(IplImage* img=0) {imgp=img;}
    ~Image(){imgp=0;}
    void operator=(IplImage* img) {imgp=img;}
    inline T* operator[](const int rowIndx) {
      return ((T *)(imgp->imageData + rowIndx*imgp->widthStep));}
  };

  typedef struct{
    unsigned char b,g,r;
  } RgbPixel;

  typedef struct{
    float b,g,r;
  } RgbPixelFloat;

  typedef Image<RgbPixel>       RgbImage;
  typedef Image<RgbPixelFloat>  RgbImageFloat;
  typedef Image<unsigned char>  BwImage;
  typedef Image<float>          BwImageFloat;

对于多通道浮动图片:

IplImage* img=cvCreateImage(cvSize(640,480),IPL_DEPTH_32F,3);
RgbImageFloat imgA(img);
imgA[i][j].b = 111;
imgA[i][j].g = 111;
imgA[i][j].r = 111;

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