C / C++ / MFC

25-Apr-12 4:01

Dear all everyone,

Please help me calling a dialog in project, I am using visual studio 2005(MFC, detail as below:
- My project have 2 dialogs(1 dialog to show data, 1 dialog login( user name and password) to protect data dialog)
- I created 1 dialog to show data is ok, but when I add a new dialog into this project and can not call dialog). So please help me!

Thanks and Best regards,

Albert Holguin25-Apr-12 4:15

Albert Holguin

25-Apr-12 4:15

How are you trying to call the dialog? ...you have to show some applicable code.

JohnCz29-Apr-12 4:54

JohnCz

29-Apr-12 4:54

I am not sure why check for valid password in OnInitDlg, while it seems that it should be checked before main dialog is invoked. No need for posting any messages to the thread.

Besides, log in dialog should not be checked for cancel only but validate password and/or username.
This code snippet shows how this can be done:

C++

	CLoginDlg loginDlg;
	loginDlg.DoModal();
	INT_PTR nResponse = loginDlg.DoModal();

	if(nResponse == IDCANCEL)
	{
		return FALSE; // canceled exit app, terminate
	}
	
	if(loginDlg.m_csUser.CompareNoCase(lpszUser))
	{
		return FALSE; // no match exit app, terminate
	}

	if(loginDlg.m_csPassword.Compare(lpszPassword))
	{
		return FALSE; // no match exit app, terminate
	}

	CMainDlg dlg;
	m_pMainWnd = &dlg;
	INT_PTR nResponse = dlg.DoModal();
.
.
.

JohnCz

Maximilien25-Apr-12 5:28

Maximilien

25-Apr-12 5:28

I would do something like :

pre-requisite.
CLoginDialog: your login dialog.
CDataDialog; your main application dialog

The CDataDialog is the main application dialog (for example created via the wizard).

When the CDataDialog is created, you can call/create the login dialog in the OnInitDialog; if the login is valid, just continue, if it is invalid, you can either close the main application, or better continue but disable some stuff or not load the data.

For example ...

C++

BOOL Ctest_dialogDlg::OnInitDialog()
{
	CDialog::OnInitDialog();

/// ....
	LoginDialog dlg;
	if ( dlg.DoModal() == IDCANCEL)
	{
		PostQuitMessage(42);
		return FALSE;
	}
	return TRUE;  // return TRUE  unless you set the focus to a control
}

Watched code never compiles.

Chris Meech25-Apr-12 7:30

Chris Meech

25-Apr-12 7:30

Maximilien wrote:
PostQuitMessage(42);

Best way to end an application ever. Smile | :)

Chris Meech
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Le Quang Long27-Apr-12 15:17

27-Apr-12 15:17

Thank you so much!
Please give me your email!

Le Quang Long28-Apr-12 18:10

28-Apr-12 18:10

Dear Maximilien,

Thank for your feedback! I want to show CLoginDialog(first). After input userID and password successfully, conntinue show CDataDialog(second) and I can type data on this CDataDialog.

Best Regards,

LQL

David Crow25-Apr-12 5:40

David Crow

25-Apr-12 5:40

Le Quang Long wrote:
...but when I add a new dialog into this project...

Are you talking about adding a new dialog template in the project's .rc file, or adding a new CDialog-based class?

Le Quang Long wrote:
...and can not call dialog).

Why not? Isn't it just like the other (first) dialog?

"One man's wage rise is another man's price increase." - Harold Wilson

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Le Quang Long25-Apr-12 14:29

25-Apr-12 14:29

I talking about adding a new dialog the same dialog-based.
Thanks,

Le Quang Long25-Apr-12 14:30

25-Apr-12 14:30

I talking about adding a new dialog the same dialog-based.
I can not call this new dialog!
Thanks,

Le Quang Long25-Apr-12 14:32

25-Apr-12 14:32

I talking about adding a new dialog the same dialog-based.
I can not call this new dialog( second dialog)! Pls give me a source code(ex)?
Thanks,

why need write itself hashtable?

Richard MacCutchan25-Apr-12 21:16

Richard MacCutchan

25-Apr-12 21:16

How can we give you source code when we have no idea what your program is doing? If you cannot call a dialog in your program then you need to give a much better explanation, including showing the code that does not work, and explaining what happens when you try.

Binding 100,000 items to a list box can be just silly regardless of what pattern you are following. Jeremy Likness

yu-jian24-Apr-12 20:27

yu-jian

24-Apr-12 20:27

I found that some open source project design itself hashtable, when need write itslef hashtable?
It will resolve which problem?

Re: why need write itself hashtable?

Richard MacCutchan24-Apr-12 22:34

Richard MacCutchan

24-Apr-12 22:34

yu-jian wrote:
It will resolve which problem?

What problem is the open source project addressing?

Binding 100,000 items to a list box can be just silly regardless of what pattern you are following. Jeremy Likness

Selecting median

a1_shay24-Apr-12 3:12

24-Apr-12 3:12

how can i Select median of two sorted arrays that not have the same size ? and also i need to find with o(logn) time.

Aescleal24-Apr-12 4:10

Aescleal

24-Apr-12 4:10

As in the median of the data set in the union of both arrays?

a1_shay24-Apr-12 4:32

24-Apr-12 4:32

the median from the both arrays together

Aescleal24-Apr-12 5:48

Aescleal

24-Apr-12 5:48

Have a look on Google for "median value of two sorted lists." There's a good algorithm on the first page.

Cheers,

Ash

a1_shay24-Apr-12 10:12

24-Apr-12 10:12

i already see it .i found this http://blogs.msdn.com/b/bali_msft/archive/2008/09/22/select-median-of-two-sorted-array.aspx[^]

look the "step 2" maybe it's my english or something else ,but i didn't understand his idea

Stefan_Lang27-Apr-12 6:54

Stefan_Lang

27-Apr-12 6:54

The idea of step 2 is this:
assuming you have two arrays with N and M elements, respectively, where N < M. Now extend that array by attaching M-N additional numbers. Then you have two arrays of the same size and can use step 1 to solve the problem.

That said, you should not actually extend the smaller array, as that would be O(N). You only have to modify the algorithm to behave as if it were extended. Also the description isn't quite correct as it fails to mention that you have to attach a like number of smaller and bigger elements to make sure the median isn't moved: so add (M-N)/2 elements that are smaller than the smallest element of both arrays at the start of the smaller array, and then (M-N)/2 elements at the end that are bigger than the biggest number in both arrays. (As I said, you don't actually add these values, you only do this virtually).

Example:

arr1 = { 6, 11, 18 } (length 3)
arr2 = { 4, 5, 8, 9, 10, 12, 15 } (length 7)

step 2 (adding 2 values at both the start and the end):
arr1a = { 1, 2, 6, 11, 18, 19, 20 } (length 7)

Actually "step 2" is a misnomer as it is supposed to be executed before step 1 in case the precondition (same array size) isn't met.

David Crow24-Apr-12 6:01

David Crow

24-Apr-12 6:01

What do two arrays have to do with the problem? Put them together into a single array, which need not be sorted. There are no fewer than five such algorithms available online (e.g., Dromey, Knuth, Floyd). This very subject was my thesis paper in graduate school.

a1_shay24-Apr-12 10:13

24-Apr-12 10:13

if i Put them together into a single array ,it's not o(logn)

David Crow24-Apr-12 17:48

David Crow

24-Apr-12 17:48

If you merged one sorted array into the other, the cost would simply be O(N log(N)). The cost for picking the element at position N/2 would then be O(1).

a1_shay25-Apr-12 4:57