Question

I’m using OpenCV within a DLL that provides plain C interfaces, no C++objects are allowed to be handed over to the calling application.

One part of this DLL performs fiducial learning for later pattern recognition which results in a list of keypoints and a Mat object. These data have to be stored by the calling application.

Handing over the keypoints via DLL interface is no problem by using a plain C struct, the members of such a keypoint can be converted easily. But I don’t see which parts of cv::Mat are really needed. Or to be more exact: my Mat-object makes use of the member “data” which points to a memory area but I have no idea how much data are contained.

So my question: how can I convert a cv::Mat object into a plain C-style structure, how can I estimate the exact length of the data field?

Thanks!

Answer 1

The easy way is to convert cv::Mat to the classical OpenCV C structure: IplImage.

cv::Mat mat = imread(...);
IplImage img(mat); // hope it's the correct syntax...

A more detailed explanation of the Mat parameters:

• data: pointer to data
• rows, columns: …
• type() – data type:
• channels() – number of channels
• step() – stride between two consecutive rows in the image, in bytes. “Includes the gaps, if any”
• size_t elemSize() similar to CV_ELEM_SIZE(cvmat->type)
• size_t elemSize1() returns the size of element channel in bytes.

And here’s how you calculate data field length:

Mat::rows * Mat::step()