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I have a program that I'd to run when the computer starts. I've put its path inside "SOFTWARE\Microsoft\Windows\CurrentVersion\Run". This is in Windows 7. When the computer starts nothing happens. I'm thinking this is because the program needs elevation when I run it. But Windows does not ask for permission to elevate and gives no feedback. It simply ignores it. I've read that Vista tells you that the program was blocked etc. Does anybody have any idea why Windows 7 simply ignores the application?

Thank you very much in advance.

Alireza

user243618
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    Because bugging the user with UAC alerts every time they turn on the computer is a bad idea. – Anon. Feb 19 '10 at 02:02
  • Out of curiosity, what is the program? – Jay Wick May 04 '10 at 04:07
  • Just a thought... but I had a similar problem because the path to my executable had slashes (/) instead of backslashes (\). I know it's dumb, but I had that issue. –  Oct 11 '10 at 11:04

1 Answers1

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Windows Vista and 7 block programs requiring elevation from startup to prevent a flurry of UAC prompts at each startup. This blog entry from the UAC team explains it pretty clearly. As a workaround, create a task scheduler entry to start the program if your program absolutely must run with elevated privileges.

Mike
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