Question

DialogBox() in WndProc()?

Asked by: bgerv

I would like my dialog box to display another dialog when a message is received. What is the effect of the following code on the calling thread? Is it allowed? Will it kill the message handling of the thread managing the MyDlgClass dialog? Should I have bidirectional messages instead (and manage the first dialog with CreateDialog)?


void MyShowDialog()
{
    .
    .
    DialogBox(hInstance, "MyDlgClass",  NULL, MyDlgProc);
    .
    .
}

INT_PTR CALLBACK MyDlgProc(HWND hWnd, UINT uMsg, WPARAM wParam, LPARAM lParam)
{
    .
    case WM_INITDIALOG:
        DialogBox(hInstance, "MyOtherDlgClass", NULL, MyOtherDlgProc);
        break;
    .
    .
}

Thanks.

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Asked On
2003-08-04 at 06:18:15ID20698826
Tags

mydlgproc

,

wndproc

,

dialogbox

,

handling

Topic

Windows Programming

Participating Experts
2
Points
50
Comments
8

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Answers

 

by: _Rob_Posted on 2003-08-04 at 06:28:47ID: 9069906

It looks like the second dialog is going to be modal with the first. Is that what you want? Or should the user be able to work with both dialogs at once?

/rob

 

by: bgervPosted on 2003-08-04 at 06:47:58ID: 9070136

The second dialog is modal with the first.

 

by: AlexFMPosted on 2003-08-04 at 07:14:17ID: 9070459

It's OK, you can do this. Internally Windows creates message loop for each modal dialog. When you call second modal dialog, Windows disables parent dialog, but it's message loop is active, so parent dialog can redraw itself. Child dialog runs with it's own message loop, when it is closed, parent is activated.

 

by: bgervPosted on 2003-08-04 at 07:23:55ID: 9070584

So you mean that the message loop starts MyDlgProc() in a new process (thread) for every message? Otherwise, MyDlgProc() (and the message loop) will be stuck in the second DialogBox() as long as the second dialog will be visible... What do you think?

 

by: AlexFMPosted on 2003-08-04 at 07:45:19ID: 9070861

No, this is the same thread, but another message loop. When your program runs, MyDloProc function instance is loaded to the stack. If this instance is handling WM_INITDIALOG, another dialog is opened. This MyDlgProc instance remains on the stack waiting for child dialog to be closed. However, message loop continues to run, and this dialog handles other messages, for example, WM_PAINT. When dialog is redrawn, another instance of MyDlgProc is loaded onto the stack, handles WM_PAINT message and unloaded.
This looks tricky because we don'r see the whole picture, we see only our message handlers called from somewhere. Adding some TRACE messages allows to understand this better.

 

by: bgervPosted on 2003-08-04 at 07:58:48ID: 9071042

I understand; thank you very much! I awarded you the points. But there is still a small point I would like to have info on. Is there something special I must do if I "block" the handler for WM_INITDIALOG? As I see it, since the first dialog is not initialized (the handler did not return), the dialog will not be shown... Or is it simply irrelevent since the return value of WM_INITDIALOG only affects the keyboard focus, which will be disabled as long as the second modal dialog will be visible?

Thanks again.

 

by: _Rob_Posted on 2003-08-04 at 08:04:04ID: 9071127

One way would be to post a user message to your window when handling the WM_INITDIALOG. Then in the handling fo your user message you display the dialog. Then the whole init process would be finished.

/rob

 

by: AlexFMPosted on 2003-08-04 at 08:07:55ID: 9071191

You don't need to do something special. However, if you want to show this dialog before another dialog is shown, post user-defined message to dialog from WM_INITDIALOG handler and show second dialog in the message handler.

20120131-EE-VQP-002

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