I am planning to use my RPi only for one application, which needs X server to be started.

This application is chromium, which requires A LOT of resources already.

I would like to start only the bare minimum (X server) to display chromium in --kiosk mode (fullscreen, can't exit, don't show desktop or anything else)

Is it possible, and if so, how?

  • Would iceweasel (firefox in disguise) be acceptable instead of Chromium which trades memory for speed? – Thorbjørn Ravn Andersen Sep 6 '17 at 16:51

You can apply hildred's suggestion permanently by creating a ~/.xinitrc file:

#!/bin/sh

exec chromium --kiosk

This will then be applied if you are using a graphical login, so that you do not have to boot to console. To test it from the console, try startx with no arguments.

  • 1
    to be more frugal with your memory use exec on the last line of your .xinitrc so you don't have a extra shell lying around using memory/swap – hildred Nov 23 '13 at 23:29
  • 1
    @hildred - just curious, approximately how much memory does a shell require? – cwd Nov 25 '14 at 4:41
  • 1
    @cwd You can get an idea with ps -o pid,rss,cmd -C bash. The RSS is in kB, it's probably 3-4000, but if you then look at top -p [pid] where pid is one of the ones reported by ps, you'll probably see 75% of that is shared (RSS vs SHR). So an extra shell takes a MB or two of RAM. I've edited hildred's suggestion into the answer. – goldilocks Nov 25 '14 at 15:17
startx chromium --kiosk --

startx is a wrapper for xinit which starts an xserver and one client program. It should do exactly what you want.

  • [1:1:13070520614:ERROR:nss_util.cc(692)] Failed to load NSS libraries. (chromium:3191): Gtk-WARNING **: cannot open display: – DrakaSAN Nov 27 '13 at 14:49

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.