How/Why is imagedata stored as char - OpenCV

I'm a tad confused.

I am just getting started with OpenCV and its image data is pointed to by a char pointer. I can't quite work out how that works considering the actual data itself could be any number of data types, e.g. uint, float, double. As far as I knew, a pointer had to be of the same type as the pointer it represents.

It's probably w开发者_Python百科orth noting that openCV is a C library and my background is C++, so I am unaware of how these problems of needing variable types are solved in C.

For example the follwing code taken from Learning OpenCV illustrates my confusion:

void saturate_sv( IplImage* img ) {
    for( int y=0; y<img->height; y++ ) {
    uchar* ptr = (uchar*) (
    img->imageData + y * img->widthStep
    );
       for( int x=0; x<img->width; x++ ) {
           ptr[3*x+1] = 255;
           ptr[3*x+2] = 255;
       }
    }
}

So this works, but when I try to operate on a iplImage of type IPL_DEPTH_64F and use ptr[3*x+1] = 1 The results are incorrect. So to distil my problems: how can I work on integer or floating point data through char pointers and specifically how could I rectify the above example to work with a double precision data.

Thanks

1. IPL_DEPTH_64F or double images will take care of the data from 0 - 1.
2. If you're used to C++ you should check out OpenCV2.0 which has several C++ classes and most importantly, one class, i.e. Mat to handle images, matrices, etc.

Here's a simple way to access elements in your image efficiently:

IplImage* img = cvCreateImage(cvSize(300,300),IPL_DEPTH_64F,1);
for( int y=0; y<img->height; y++ ) 
    {
       double* ptr = reinterpret_cast<double*>(img->imageData + y * img->widthStep);
       for( int x=0; x<img->width; x++ ) 
       {
          ptr[x] = double(255);
       }
    }
cvNamedWindow("SO");
cvShowImage("SO",img);
cvWaitKey();
cvDestroyAllWindows();
cvReleaseImage(&img);

Since you're working with a double image, it makes more sense to:

1. Work with a double pointer so you can easily assign elements in a row with ptr[x]
2. Do the pointer arithmetic in bytes (img->imageData + y * img->widthStep) and the cast it to a double pointer

Also, it's important that you do the pointer arithmetic in bytes (or uchar, i.e. unsigned char) since OpenCV tends to pad the rows of the images with extra bytes for efficiency (especially for double images).

So even if a double element is 8 bytes, and you have, say, 300 rows, a row is not guaranteed to end at 8*300 or 2400 bytes since OpenCV might pad the end.

Therefore, this prevents you from initializing a pointer to the first element of the image and then using ptr[y*img->height+x] to access elements since each row might have more than 8*(y*img->height) bytes.

That's why the example code calculates the pointer to each row each time using img->widthStep which represents the true size of each row in bytes.

OpenCV 2.0

If you use the Mat class, you can do the same thing along these lines:

cv::Mat img(300,300,CV_64FC1);
for( int y=0; y<img.rows; y++ ) 
    {
       double* ptr = reinterpret_cast<double*>(img.data + y * img.step);
       for( int x=0; x<img.cols; x++ ) 
       {
          ptr[x] = double(255);
       } 
    }
cv::namedWindow("SO");
cv::imshow("SO",img);
cv::waitKey();

where img.step is the distance between successive rows in bytes

And if you want to directly access the element (slower):

img.at<double>(y,x)