December, 2004

  • The Old New Thing

    Dragging a shell object, part 1: Getting the IDataObject

    • 20 Comments

    The shell gives you the IDataObject; all you have to do is drag it around. (This is the first of a five-part series.)

    Start with the scratch program, and add the function GetUIObjectOfFile from an earlier article. Also, change the calls to CoInitialize and CoUninitialize to OleInitialize and OleUninitialize, respectively, since we're now going to be using full-on OLE and not just COM.

    In order to initiate a drag/drop operation, we need a drop source:

    class CDropSource : public IDropSource
    {
    public:
      // *** IUnknown ***
      STDMETHODIMP QueryInterface(REFIID riid, void **ppv);
      STDMETHODIMP_(ULONG) AddRef();
      STDMETHODIMP_(ULONG) Release();
    
      // *** IDropSource ***
      STDMETHODIMP QueryContinueDrag(BOOL fEscapePressed, DWORD grfKeyState);
      STDMETHODIMP GiveFeedback(DWORD dwEffect);
    
      CDropSource() : m_cRef(1) { }
    private:
      ULONG m_cRef;
    };
    
    HRESULT CDropSource::QueryInterface(REFIID riid, void **ppv)
    {
      IUnknown *punk = NULL;
      if (riid == IID_IUnknown) {
        punk = static_cast<IUnknown*>(this);
      } else if (riid == IID_IDropSource) {
        punk = static_cast<IDropSource*>(this);
      }
    
      *ppv = punk;
      if (punk) {
        punk->AddRef();
        return S_OK;
      } else {
        return E_NOINTERFACE;
      }
    }
    
    ULONG CDropSource::AddRef()
    {
      return ++m_cRef;
    }
    
    ULONG CDropSource::Release()
    {
      ULONG cRef = --m_cRef;
      if (cRef == 0) delete this;
      return cRef;
    }
    
    HRESULT CDropSource::QueryContinueDrag(
              BOOL fEscapePressed, DWORD grfKeyState)
    {
      if (fEscapePressed) return DRAGDROP_S_CANCEL;
    
      // [Update: missing paren repaired, 7 Dec]
      if (!(grfKeyState & (MK_LBUTTON | MK_RBUTTON)))
        return DRAGDROP_S_DROP;
    
      return S_OK;
    }
    
    HRESULT CDropSource::GiveFeedback(DWORD dwEffect)
    {
      return DRAGDROP_S_USEDEFAULTCURSORS;
    }
    

    As you can see, this drop source is extraordinarily boring. Even the interesting methods are uninteresting.

    The IDropSource::QueryContinueDrag method is pretty much boilerplate. If the Escape key was pressed, then cancel the drag/drop operation. If the mouse buttons are released, then complete the operation. Otherwise, continue the operation.

    The IDropSource::GiveFeedback method is even less interesting. It merely returns DRAGDROP_S_USEDEFAULTCURSORS to indicate that it wants default drag feedback.

    Believe it or not, we now have everything we need to drag a file.

    void OnLButtonDown(HWND hwnd, BOOL fDoubleClick,
                       int x, int y, UINT keyFlags)
    {
      IDataObject *pdto;
      // In a real program of course
      // you wouldn't use a hard-coded path.
      // [comment added 11am because apparently some
      // people thought this wasn't self-evident.]
      if (SUCCEEDED(GetUIObjectOfFile(hwnd,
                        L"C:\\Windows\\clock.avi",
    		    IID_IDataObject, (void**)&pdto))) {
        IDropSource *pds = new CDropSource();
        if (pds) {
          DWORD dwEffect;
          DoDragDrop(pdto, pds, DROPEFFECT_COPY | DROPEFFECT_LINK,
                     &dwEffect);
          pds->Release();
        }
        pdto->Release();
      }
    }
    
        HANDLE_MSG(hwnd, WM_LBUTTONDOWN, OnLButtonDown);
    

    To drag an object, you need two things, a data object and a drop source. We created our drop source above, and the data object comes from the shell. All that's left to do is start the drag/drop operation by calling the DoDragDrop function.

    Notice that we specify that the permitted operations are DROPEFFECT_COPY and DROPEFFECT_LINK. We specifically disallow DROPEFFECT_MOVE because this program doesn't present a folder-like window; the user has no expectation that the drag/drop will result in a Move operation.

    Next time, adding Move support, just to see how it works.

  • The Old New Thing

    Why are documents printed out of order when you multiselect and choose "Print"?

    • 26 Comments

    If you select say five files and then right-click them and choose "Print", they tend to print in a random order. Why is that?

    The shell invokes the Print verb on each file in turn, and depending on how the program responsible for printing the document is registered, one of several things can happen.

    • Most commonly, the program that prints the document registered a simple command line under the shell\print\command registry key. In this case, the program is launched five times, each with a different file. All these print commands are now racing to the printer and it's a question of which copy of the program submits its print job first that determines the order in which they come out of the printer. (You're probably going to see the shortest and simplest documents come out first since they take less time to render.)
    • Occasionally, the program that prints the document registered a DDE verb under the shell\print\ddeexec registry key. In this case, one copy of the program is launched and it is given each filename one at a time. What it does with those filenames is now up to the program. If the program supports background printing, it will probably shunt the printing of the document onto a background thread, and now you're roughly in the same fix as the previous scenario: Five background threads each racing to see who can submit their print job first.
    • Extremely rarely, the program that prints the document registered a drop handler under the shell\print\DropTarget key. In this case, the drop target is instantiated and is given the list of files. It is then up to the drop target to decide what to do with the documents.

    These three ways of registering print actions are described in the MSDN documentation on verbs and file associations.

    [Update: 7:30am, fixed title.]

  • The Old New Thing

    Why did Windows 95 run the timer at 55ms?

    • 32 Comments

    The story behind the 55ms timer tick rate goes all the way back to the original IBM PC BIOS. The original IBM PC used a 1.19MHz crystal, and 65536 cycles at 1.19MHz equals approximately 55ms. (More accurately, it was more like 1.19318MHz and 54.92ms.)

    But that just pushes the question to another level. Why 1.19...MHz, then?

    With that clock rate, 216 ticks equals approximately 3600 seconds, which is one hour. (If you do the math it's more like 3599.59 seconds.) [Update: 4pm, change 232 to 216; what was I thinking?]

    What's so special about one hour?

    The BIOS checked once an hour to see whether the clock has crossed midnight. When it did, it needed to increment the date. Making the hourly check happen precisely when a 16-bit tick count overflowed saved a few valuable bytes in the BIOS.

    Another reason for the 1.19MHz clock speed was that it was exactly one quarter of the original CPU speed, namely 4.77MHz, which was in turn 4/3 times the NTSC color burst frequency of 3.5MHz. Recall that back in these days, personal computers sent their video output to a television set. Monitors were for the rich kids. Using a timer related to the video output signal saved a few dollars on the motherboard.

    Calvin Hsia has another view of the story behind the 4.77MHz clock.

    (Penny-pinching was very common at this time. The Apple ][ had its own share of penny-saving hijinks.)

  • The Old New Thing

    What is the purpose of the bmPlanes member of the BITMAP structure?

    • 21 Comments

    Many bitmap-related structures in Windows have a field called "planes". For example the BITMAPINFOHEADER structure has a biPlanes member (which must be set to 1). The BITMAP structure has a field called bmPlanes. What's the deal with that field?

    The EGA video adapter supported 16 simultaneous colors. This was an enormous improvement over the CGA, which supported only four colors. If you have 16 colors, then you need four bits per pixel. You would think that the encoding would be to have the each byte of video memory encode two pixels, one in the bottom four bits and one in the top four. But for technical reasons, the structure of video memory was not that simple.

    Instead of putting the bits for a single pixel next to each other, the color channels were each split into their own monochrome bitmap. In other words, the pixels were sliced "the other way":

    0 0 0 0 1 1 1 1 | 0F
    0 0 1 1 0 0 1 1 | 33
    0 1 0 1 0 1 0 1 | 55
    0 1 1 0 0 1 1 0 | 66

    0 3 5 6 8 B D E

    Suppose you wanted to display eight pixels, with colors { 0, 3, 5, 6, 8 B, D, E } above. Instead of storing the nibbles in that order, slice the nibbles apart into their component bits and collect all the bits from the same position together. In other words, read the bits across rather than down.

    In the default 16-color palette, the colors were assigned so that bit 0 was the blue channel, bit 1 was the green channel, bit 2 was the red channel, and bit 3 was the intensity channel. With this interpretation, the four slices can be interpreted as the "intensity plane", the "red plane", the "green plane" and the "blue plane". For the last three planes, you can imagine that each one represents what you would see if only the corresponding electron gun were firing.

    Since this was the native color format for EGA, there needed to be a way to express this color format in the BITMAP structure so that device-dependent bitmaps could be represented by Windows.

    Thus was born the planar color format. For 16-color planar bitmaps, the number of planes is four and the number of bits per pixel is one.

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