Conversion from MFC CImage to Opencv IPlImage - c++

Question

I am having a camera application using opencv where I am using MFC GUI.I need to convert the CImage to IplImage for the Opencv modules for image processing and convert it back to CImage for displaying in the window again.
I researched in this topic but not enough examples and solutions.Anybody have some suggestions.Thanks
This is my code...
void CChildView::OnFileOpenimage(void)
{
// TODO: Add your command handler code here
CString strFilter;
CSimpleArray<GUID> aguidFileTypes;
HRESULT hResult;
hResult = imgOriginal.GetExporterFilterString(strFilter,aguidFileTypes);
if (FAILED(hResult)) {
CString fmt;
fmt.Format("GetExporterFilter failed:\n%x - %s", hResult, _com_error(hResult).ErrorMessage());
::AfxMessageBox(fmt);
return;
}
CFileDialog dlg(TRUE, NULL, NULL, OFN_FILEMUSTEXIST, strFilter);
dlg.m_ofn.nFilterIndex = m_nFilterLoad;
hResult = (int)dlg.DoModal();
if(FAILED(hResult)) {
return;
}
m_nFilterLoad = dlg.m_ofn.nFilterIndex;
imgOriginal.Destroy();
CString pathval = dlg.GetPathName();
hResult = imgOriginal.Load(dlg.GetPathName());
if (FAILED(hResult)) {
CString fmt;
fmt.Format("Load image failed:\n%x - %s", hResult, _com_error(hResult).ErrorMessage());
::AfxMessageBox(fmt);
return;
}
// IplImage *img from imgOriginal; want to convert here for further processing.
m_nImageSize = SIZE_ORIGINAL;
Invalidate();
UpdateWindow();
}

Answer 1

Had this problem for quite a while myself. I found this code here which I ended up using that worked quite well. Instead of researching "CImage to IplImage" you should search "HBITMAP to IplImage"
IplImage* hBitmap2Ipl(HBITMAP hBmp, bool flip)
{
BITMAP bmp;
::GetObject(hBmp,sizeof(BITMAP),&bmp);
int nChannels = bmp.bmBitsPixel == 1 ? 1 : bmp.bmBitsPixel/8;
int depth = bmp.bmBitsPixel == 1 ? IPL_DEPTH_1U : IPL_DEPTH_8U;
IplImage *img=cvCreateImageHeader(cvSize(bmp.bmWidth, bmp.bmHeight), depth, nChannels);
img->imageData = (char*)malloc(bmp.bmHeight*bmp.bmWidth*nChannels*sizeof(char));
memcpy(img->imageData,(char*)(bmp.bmBits),bmp.bmHeight*bmp.bmWidth*nChannels);
return img;
}
Usage:
ATL::CImage image;
//whatever code you use to generate the image goes here
IplImage *convertedImage=hBitmap2Ipl(image.Detach());
image.Destroy();

Answer 2

I searched half the night for my copy-and-paste memory leak and finally found it!
So here is an updated code snippet of the original answer. The important thing for me was to use cvCreateImage instead of cvCreateImageHeader because I did use cvReleaseImage after the call to hBitmap2IplImage. Also flipping was neccessary in my case. Cheers!
IplImage* hBitmap2IplImage(HBITMAP hBmp, bool flip)
{
BITMAP bmp;
::GetObject(hBmp,sizeof(BITMAP),&bmp);
int nChannels = bmp.bmBitsPixel == 1 ? 1 : bmp.bmBitsPixel/8;
int depth = bmp.bmBitsPixel == 1 ? IPL_DEPTH_1U : IPL_DEPTH_8U;
IplImage *img=cvCreateImage(cvSize(bmp.bmWidth, bmp.bmHeight), depth, nChannels);
size_t imgSize = bmp.bmHeight*bmp.bmWidth*nChannels;
memcpy_s(img->imageData, imgSize, (char*)(bmp.bmBits), imgSize);
if (flip)
cvFlip(img, NULL, 0);
return img;
}
Usage:
ATL::CImage image;
HBITMAP hBitmap = image.Detach(); // some handle to your bitmap
bool flip = true;
IplImage* pSomeIplImage = hBitmap2IplImage(hBitmap, flip);
image.Destroy();
//
// use "pSomeIplImage"
//
cvReleaseImage( &pSomeIplImage ); // don't forget

Answer 3

If a bitmap image is on a view you can use the device context(CDC dcBuffer ) to copy it.
IplImage *pCapture = cvCreateImage(cvSize(U16_IMAGE_WIDTH/2,U16_IMAGE_HEIGHT/2),8, 3);
::GetDIBits(dcBuffer.GetSafeHdc(), bitmapBuffer, 0, U16_IMAGE_HEIGHT/2, pCapture->imageData, &bitmapInfo, DIB_RGB_COLORS);
cvSaveImage("temp.bmp", pCapture);
cvReleaseImage(&pCapture);