Question

I am trying to start a program I made in this directory:

C:\example\example.exe -someargument

when the computer starts up. I am attempting to use this registry key:

HKEY_LOCAL_MACHINE\Software\Microsoft\Windows\CurrentVersion\Run

with the key being:

Name: example
Type: REG_SZ
Data: "C:\example\example.exe -someargument"

But my program also needs files from the directory C:\example but can’t find them since the current working directory is different. Is is possible to do something like this in the registry key value

"cd C:\example\; example.exe -someargument"

so that it will change the directory? Or is there a better solution?

Thanks!

Answer 1

You can register your application under next registry key (like this does Reg2Run tool)

HKEY_CURRENT_USER\Software\Microsoft\Windows\CurrentVersion\App Paths\example.exe

@="c:\example\example.exe"
Path="c:\AnotherPath"

So System.Diagnostics.Run("example.exe"); will launch your application with specified working path.

Or another way: write a launcher using C#. You can do the same using a PowerShell cmdlet.

var info = new System.Diagnostics.ProcessStartInfo(@"c:\example\example.exe", "-someargument")
{
    WorkingDirectory = @"c:\AnotherPath"
};
System.Diagnostics.Process.Start(info);