Why CMFCPopupMenu avoid submeu's uistate?

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//Visual Studio 2008 sp 1

when I use GetContextMenuManager()->ShowPopupMenu(), it avoid submenu's uistate.

//ShowPopupMenu(HMENU hmenuPopup, int x, int y, CWnd* pWndOwner, BOOL bOwnMessage = FALSE, BOOL bAutoDestroy = TRUE, BOOL bRightAlign = FALSE);

Example source code

CMenu menu1, menu2;
menu.CreatePopupMenu();

menu1.AppendMenu( MF_STRING | MF_CHECKED, WM_USER+1, _T("A") );
menu2.AppendMenu( MF_STRING | MF_CHECKED, WM_USER+2, _T("SubMenu A") );

menu1.AppendMenu( MF_BYCOMMAND | MF_STRING | MF_POPUP, reinterpret_cast<uint_ptr>(menu2.Detach()), _T("Submenu") );

theApp.GetContextMenuManager()->ShowPopupMenu(menu1.Detach(), x, y, this, TRUE, TRUE);

when it appeared, the menu "A" is checked, but SubMenu A wasn't.

isn't there anyone know why?

Posted 15-Feb-10 21:30pm

Ummae

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2 solutions

Solution 2

//Eugen Podsypalnikov
Thanks. : )

In fact, i know it there no the process that for submenu's uistate.
Actually, what i want to know is why they made it like this? I fell that it's some kind of bug..

Posted 15-Feb-10 23:42pm

Ummae

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Eugen Podsypalnikov · Accepted Answer · 2010-02-15T22:30:00

Solution 1

The function

C++

void CMFCToolBarMenuButton::CreateFromMenu(HMENU hMenu)

does produce a submenu, but does not control the checked state inside...

You could insert your buttons directly (it's easy),
or reaccess some buttons after the creating of presented kind :)

Posted 15-Feb-10 22:30pm

Eugen Podsypalnikov