I want my application to start at windows startup BUT with a small timeout

[Leon Vitanos]

so i want for example to wait 20 seconds and then start the app

how should my registry at HKEY_CURRENT_USER\\Software\\Microsoft\\Windows\\CurrentVersion\\Run
be changed?

instead of the path of the exe can i have somehow, a timeout and then the path of the exe?

i know there exists one command at cmd --timeout, but how i can implement it at startup?

[kevinkjt2000]

There are several examples given here on the windows 7 forums:
http://www.sevenforums.com/general-discussion/89611-anyway-delay-start-up-program-20-sec-more.html

The link from there that concerns regedit:
http://support.microsoft.com/kb/193888

Basically if your app has a service, you can make that service dependent on a network/internet service.

[modoran]

Change the app itself, add a Sleep() for 20 seconds or so in it. No service required.

If that is not possible just write another small app that wait 20 seconds, launch the main app and exits.

[Leon Vitanos]

Yes my application is not registered as a service, so i guess i have to go with the sleep way at my application's code.

But just so i learn. How do you register ur app as a service? Is it a registry? Is it necessary/better?

Thanks

[kevinkjt2000]

I believe I found an example written in VC++ here:
http://code.msdn.microsoft.com/windowsdesktop/CppWindowsService-cacf4948