“Start up” launching two instances of same program

Question

I wrote a simple Windows program and put it in my "Start up" folder on the start menu.

But when Windows starts up, it always launches it twice. I get two instances of my program and I always have to manually exit one of them.

It is only in the All Users start up profile, not any of the other User profiles. I haven't touched the registry at all. If I remove the program from "Start up", then it doesn't start at all.

This actually happens on both 7 and xp for me for this particular program.

Answer 1

Go to the run dialog box and type in msconfig.

A box like the one below should come up:

Have a look on the startup tab and uncheck the entries for the application that is launching twice.

Answer 2

Check that the program isn't installed in "All Users\Starup" as well as "Your Username\Startup"

This happened to me once when I was fiddling with the start menu.

Answer 3

You Could make try making this registry file (remove the shortcut from all users start-up first)

----------------------------------------start copy

Windows Registry Editor Version 5.00

[HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Run]
"{Software Name}"="\"C:\\Program Files\\{Install Folder}\\{Application Name.exe}\""

end copy-----------------------------------------

Change the {folder and file names} to what ever you need to start, removing the curvy brackets.

Save it as a .txt file then change the extension to .reg

Run the file and see if it corrects the issue

If not you can always remove the entry by creating a .reg file with this line (remeber to change the {software Name} to what every you changed the first one to.

------------------------start copy

Windows Registry Editor Version 5.00

[HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\Run]
"{Software Name}"=""

end copy---------------------------