Copying OpenCV mat to a known address with memcpy

Question

I'm very new at C++ and I'm trying to create a DLL which uses OpenCV library.

My DLL gets a raw image from other application and creates a MAT from the application's memory buffer. I send the buffer's address, which has a raw image, to the DLL and get raw image to OpenCV. This part works. But after processing image with OpenCV, I can't write raw image to same memory address.

This is the code snippet that I've tried:

fn_export double createImage(char* address  double width, double height) {
  
  unsigned char* pBuffer = (unsigned char*)address;
  memcpy(&pBuffer,&address, sizeof(pBuffer));
  cv::Mat img = cv::Mat(height,width, CV_8UC4, pBuffer);
  cv::imshow("Original", img);

  memcpy(&address, &img.data[0], sizeof(address));

  return 1;
}

char* address is memory address from my application. Other application's buffer doesn't change this way. Anybody has any advice about this situation?

In memcpy() when you take sizeof(address), this is incorrect because this is just copying the pointer not the actual data at the location where the pointer points to. Also &address[0] and &address are the same thing. — , Jul 28 '20 at 10:36
Thank you for your comment. Yes, you are right. I realized that some definitions in my code are unnecessary. — Mb B, Aug 02 '20 at 18:07

score 2 · Accepted Answer · answered Jul 26 '20 at 14:29

2

Ok. I solved this issue ;

    Mat img = Mat(height, width, CV_8UC4, address);
cv::imshow("Image from GM", img);
// same image copy to buffer back;
memcpy(&address[0], &img.data[0], width*height*4.);

answered Jul 26 '20 at 14:29

Mb B

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Technically, you should check if the image is contiguous before assuming it is and copying without checking... https://stackoverflow.com/q/33665241/2836621 – Mark Setchell Jul 26 '20 at 17:44
Mark the problem as answered. – Jul 28 '20 at 10:37

Copying OpenCV mat to a known address with memcpy

1 Answers1