How to make BitBlt() draw in right-to-left orientation?

learner 0 Reputation points
2024-10-04T11:08:25.4166667+00:00

Hello,

I'm using BitBlt() to paint a horizontal bar on top of a splitter in my app, which supports languages that follow left-to-right orientation (English), as well as, right-to-left orientation (Arabic).

BitBlt() seems to pretty much follow left-to-right orientation and when we pass X and Y-coordinate, it draws to the right of the coordinates. Now, I want the opposite for collation that follows right-to-left. I have the coodinates and I just want BitBlt() to draw to the left of the coodinates this time.

Is it possible?

Code -

BitBlt(destination_handle, X, Y, length_of_black_bar, width_of_black_bar, 0, 0, 0, DSTINVERT);

Thanks.

Windows API - Win32
Windows API - Win32
A core set of Windows application programming interfaces (APIs) for desktop and server applications. Previously known as Win32 API.
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  1. Castorix31 85,956 Reputation points
    2024-10-04T12:59:41.6966667+00:00

    If you want to mirror it from right to left, a way is with StretchBlt

    For example :

    					HBITMAP hBitmap = (HBITMAP)LoadImage(NULL, TEXT("E:\\takingphoto.bmp"), IMAGE_BITMAP, 0, 0, LR_LOADFROMFILE);
    
    					int nX = 200, nY = 100;
    					BITMAP bm;
    					GetObject(hBitmap, sizeof(BITMAP), &bm);
    
    					HDC hDC = GetDC(NULL);
    					HDC hDCMem = CreateCompatibleDC(hDC);
    					HBITMAP hBitmapOld = (HBITMAP)SelectObject(hDCMem, hBitmap);
    					BitBlt(hDC, nX, nY, bm.bmWidth, bm.bmHeight, hDCMem, 0, 0, SRCCOPY);
    					nY += bm.bmHeight + 10;
    					StretchBlt(hDC, nX, nY, bm.bmWidth, bm.bmHeight, hDCMem,
    						bm.bmWidth - 1, 0,
    						-bm.bmWidth, bm.bmHeight,
    						SRCCOPY);
    					SelectObject(hDCMem, hBitmapOld);
    					DeleteObject(hDCMem);
    					ReleaseDC(NULL, hDC);
    
    

  2. learner 0 Reputation points
    2024-10-15T17:49:41.2166667+00:00

    Hi,

    For me, providing a negative width to BitBlt() function worked. Thanks!

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