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

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?
There are several examples given here on the windows 7 forums:

The link from there that concerns regedit:

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

I believe I found an example written in VC++ here:
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