invoke ShellExecute,0,0,'rundll32.exe','Shell32.dll,Control_RunDLL Mmsys.cpl,,1',0,SW_SHOW
    

Hi
I need to open Win settings applet by shortest way.

rundll32.exe Shell32.dll,Control_RunDLL Mmsys.cpl,,0
rundll32.exe Shell32.dll,Control_RunDLL Mmsys.cpl,,1

I'm trying to do

Code:

invoke ShellExecute,0,0,'rundll32.exe','Shell32.dll,Control_RunDLL Mmsys.cpl,,1',0,SW_SHOW
    

format PE GUI 4.0
entry start

include 'win32a.inc'

section '.data' readable writable
       _caption  db 'SHELL',0
       _message  db 'Open Windows Settings?',0
       _param    db 'shell32.dll,Control_RunDLL mmsys.cpl,,0',0
       _file     db 'rundll32.exe',0

section '.code' code readable writable executable

start:
       invoke  MessageBox, 0, _message, _caption, MB_ICONQUESTION + MB_YESNO
       cmp     eax,IDYES
       jne     .exit

       invoke ShellExecute, 0, 0, _file, _param, 0, SW_SHOW

.exit:
       push eax
       call [ExitProcess]

section '.idata' import readable writable

       library kernel, 'KERNEL32.DLL',\
               user, 'USER32.DLL', \
               shell, 'SHELL32.DLL'

       import kernel,\
              ExitProcess, 'ExitProcess'

       import user,\
              MessageBox, 'MessageBoxA'

       import shell,\
              ShellExecute, 'ShellExecuteA'