setDIbitstodevice Turbo C++

Question

The program I have uses setDIbitstodevice for the camera.
I see that the code is written as:

SetDIBitsToDevice( PaintBox->Canvas->Handle, 0, 0, 160, 120,
    0, 0, 0, 120,
    Image, 
    ( LPBITMAPINFO )&mapHead, DIB_RGB_COLORS );

And:

Image = new unsigned char [ 160 * 120 * 4 ]

I do not know what does it mean. I was told that if I change Image to other form of unsigned char , the picture display would change as well but I did that and I couldn't get anything.

Please advice, thanks.

Answer 1

unsigned char has size of a byte in turbo C++, so the 'Image = new... ' line is assigning some memory to Image, which is 160 pixels wide, by 120 pixels high, and 4 bytes deep. The 4 bytes will be one byte each for red, green, blue and alpha channel (although these don't support an alpha channel so I don't think that's quite right - but it's allocating more memory than necessary, so it wont crash, but it's probably wasting memory).

Now SetDIBitsToDevice takes the pixels in the Image and displays it to the paintbox (via the handle). Tee parameters are the size of the destination and so on (see http://msdn.microsoft.com/en-us/library/dd162974(VS.85).aspx ). If you want to stretch the bitmap (ie change the displayed size) then have a look at StretchDIBits http://msdn.microsoft.com/en-us/library/dd145121(v=VS.85).aspx .

hth