How to start application at startup of booting time.

  • Thread starter Thread starter Abhijit D. Babar
  • Start date Start date
A

Abhijit D. Babar

As you say to make changes as follows on registry:-
[HKEY_LOCAL_MACHINE\Init]
Launchxx = "MyShell.EXE"
Dependsxx = hex:14, 00, 1e, 00
Click to expand...


but problem is i given below.
i have hard disk with primary partition 32mb. I have 3 files
boot.ini, splash.bmx and eboot.bix. and also nk.bin on active
partition. when i start the device, image is loaded from hard disk
to ram. then how to edit the registry, because it is loaded on ram. I
place the application exe on \windows\startup folder, but these folder
are created on ram, when i restart the target device, this file are
removed. Then how to start application at startup of booting time
 
If the registry is in RAM, then there's nothing that you can do unless the
operating system itself is built in such a way, by the hardware OEM, to
provide another means of starting an application on boot. Did you build the
operating system? If so, then set it up to use one of the persistent
registry schemes (hive or OEM -- read the Platform Builder help). If you
are not the device OEM, ask them! We can't make the impossible possible.

Paul T.

Abhijit D. Babar said:
As you say to make changes as follows on registry:-
[HKEY_LOCAL_MACHINE\Init]
Launchxx = "MyShell.EXE"
Dependsxx = hex:14, 00, 1e, 00
Click to expand...


but problem is i given below.
i have hard disk with primary partition 32mb. I have 3 files
boot.ini, splash.bmx and eboot.bix. and also nk.bin on active
partition. when i start the device, image is loaded from hard disk
to ram. then how to edit the registry, because it is loaded on ram. I
place the application exe on \windows\startup folder, but these folder
are created on ram, when i restart the target device, this file are
removed. Then how to start application at startup of booting time
Click to expand...
 
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