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I wrote a small utility tool with MFC (for user interface) and purely C++ code.
on Windows 10 and in Visual Studio 2017 IDE, I compiled this program into .exe file (i.e. HelloX.exe) in MFC dynamic DLL option.
So I put vcruntime140.dll and mfc140u.dll at the same folder.
I test my program on Windows Vista, it works well. I just tested this program on Windows XP and it did not work(few minutes ago).
Any suggestion or comments to share? thanks a million.
diligent hands rule....
modified 8-Mar-20 22:14pm.
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thanks
diligent hands rule....
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Deploy as in creating an installer ?
For small packages like yours, I would use inno setup to package your software.
I'd rather be phishing!
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not using installer. just xcopy command etc.
diligent hands rule....
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To deploy without having to install the VC++ dlls.
Linking to C runtime statically (Multithreaded (/MT) for Release build)
http://imageshack.com/a/img923/6600/ifeLcs.png
If you are using MFC, linking to MFC runtime statically.
http://imageshack.com/a/img921/1331/7eBoZW.png
The downside to linking statically is your executable size will be bigger but it only links the classes and functions your executable calls whereas, in the dynamically linkage case, the VC++ dlls have to contains all the code regardless your executable calls them or not.
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I plan to use MFC DLL dynamically linked option. I have several utilities to distribute in one bundle...
diligent hands rule....
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thank you! first to know VS 2015, VS 2017,and VS2019 have the same redistribute.
diligent hands rule....
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Yes, this is the future direction going forward because MS does not want the user to install many VC++ redistributables on their machine.
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your screenshots are very helpful. thanks.
diligent hands rule....
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hi
I am trying to add a MENU bar to my Dialog when I look at this link seems I should be able to
DIALOG resource - Win32 apps | Microsoft Docs[^]
however after I add it
IDD_DIALOG21 DIALOGEX 0, 0, 790, 321
MENU IDR_MAINFRAME
STYLE DS_SETFONT | DS_MODALFRAME | DS_FIXEDSYS | WS_POPUP | WS_CAPTION | WS_SYSMENU
CAPTION "Dialog"
FONT 8, "MS Shell Dlg", 400, 0, 0x1
BEGIN
CONTROL "",IDC_RICHEDIT21,"RichEdit20A",ES_MULTILINE | WS_BORDER | WS_TABSTOP,27,7,780,273
LTEXT "Static",IDC_STATIC15,7,7,20,273,WS_CLIPSIBLINGS
END
and open up the dialog in the resource editor it doesn't display
I tried creating a new project "Dialog Based" and the menu (control to add it ) in the wizard was grayed out
thanks
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I don't have a VS version that includes the resource editor, so I have to do things manually. However, I can confirm that dialog boxes work fine with menus.
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ok maybe it only display them when you do a SHowWindow
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I thought you were talking about a modal dialog.
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Easy to do it manually .. inside WM_INITDIALOG assuming the dialog window handle is hWnd
Try this
HMENU hMenubar;
HMENU hMenu;
hMenubar = CreateMenu();
hMenu = CreateMenu();
AppendMenuW(hMenu, MF_STRING, 10000, "&New");
AppendMenuW(hMenu, MF_STRING, 10001, "&Open");
AppendMenuW(hMenu, MF_SEPARATOR, 0, NULL);
AppendMenuW(hMenu, MF_STRING, WM_QUIT, "&Quit");
AppendMenuW(hMenubar, MF_POPUP, (UINT_PTR)hMenu, "&File");
SetMenu(hWnd, hMenubar);
In vino veritas
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ForNow wrote: after I add it
IDD_DIALOG21 DIALOGEX 0, 0, 790, 321
MENU IDR_MAINFRAME
...
END
and open up the dialog in the resource editor it doesn't display
Did you create the menu IDR_MAINFRAME itself?
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IDR_MENU
is defined in my resource file I basically took the menu for the MainFrame and wanted to use it for the Dialog
Opening up The resource editor I am beginning to think when you view a dialog with menu it wont display it
There is a only a way to view the menu by itself
once I get my code going and do a ShowWindow for the Dialog i'll see if it displays the menu
IDR_MAINFRAME MENU
BEGIN
POPUP "&File"
BEGIN
MENUITEM "E&xit", ID_APP_EXIT
END
POPUP "&Edit"
BEGIN
MENUITEM "&Undo\tCtrl+Z", ID_EDIT_UNDO
MENUITEM SEPARATOR
MENUITEM "Cu&t\tCtrl+X", ID_EDIT_CUT
MENUITEM "&Copy\tCtrl+C", ID_EDIT_COPY
MENUITEM "&Paste\tCtrl+V", ID_EDIT_PASTE
END
POPUP "&View"
BEGIN
MENUITEM "&Toolbar", ID_VIEW_TOOLBAR
MENUITEM "&Status Bar", ID_VIEW_STATUS_BAR
END
POPUP "&ProgDebug"
BEGIN
MENUITEM "Program Debug", ID_DEBUG
END
POPUP "&Asidlist"
BEGIN
MENUITEM "Address Space List", ID_ASID
END
POPUP "&Help"
BEGIN
MENUITEM "&About DBGR...", ID_APP_ABOUT
END
END
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I wanted to launch 'Caliberate the screen for Pen or touch input' (Tablet PC Settings) using Rundll32 Commands in Windows 10.
for ex like : rundll32.exe shell32.dll,Control_RunDLL tabletpc.cpl
Is anyone have an idea how to do it for this option. tabletpc.cpl option opening 'Pen and touch' window. I wanted to open Tablet PC Setting window not the Pen and Touch window.
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Hello how to create a C++ application to send a HTTP request for Dialer.exe in Win32 using TSP ?
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An attempt to learn and make sense of evaluating results.
Making these assumptions, right or wrong
"bool" is not "standard" C/C++ type
when "condition" such as in "if(condition)" evaluates to true , it is binary zero
thus if(condition == 0) would make better sense
then if(condition)
most "well written functions return x" , x being mostly zero when function is successful
when function fails – the return value is (generally) -1 or positive value identifying the error
then same as above - if(function (z) == 0) should prevail.
Of course explicit evaluation of result to zero could prevent hard to locate bugs when these commonly used implicit evaluation conventions are not followed by author of the code..
Any other views / comments would be appreciated.
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Quote: "bool" is not "standard" C/C++ type
Well... Fundamental types - cppreference.com[^], Boolean type support library - cppreference.com[^].
Quote: thus if(condition == 0) would make better sense
then if(condition)
most "well written functions return x" , x being mostly zero when function is successful
when function fails – the return value is (generally) -1 or positive value identifying the error
then same as above - if(function (z) == 0) should prevail.
In C++, if a function returns a bool then using
if ( function(x) ) is the only meaningful way I see.
On the other hand, if the function returns an integer data type, then
if (function(x) == expected_int_result) is the most sensible approach.
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