How can I start a login shell with options?

Question

I am messing around and want to have a guest user whose login shell is a restricted one like rbash. Unfortunately there is no rbash executable anywhere on my Slackware installation, so I have to do bash -r instead.

I tried putting this in /etc/passwd as the login shell for the user in question: guest:x:1001:1001::/home/guest:/bin/bash -r but it seems to be trying to find an executable with " -r" at the end of the filename and so doesn't let me log in because it doesn't exist.

Is there any way I can start a restricted shell with an option like this, or do I have to install an rbash binary? This is not ideal as I am on a Raspberry Pi (ARM) and may have to compile a package which is something I am unfamiliar with.

I don't think there is an "rbash binary" - it's just a symlink to `bash`, which uses `argv[0]` to decide whether to run in restricted mode. At least that's how it appears to work on Debian-based systems. — steeldriver, Apr 24 '21 at 13:08
Can symlinks pass options to executables they point to? If so how do you do it? — Architect, Apr 24 '21 at 13:10
Ah, I see. I'll write an answer explaining exactly what I did. — Architect, Apr 24 '21 at 13:18

score 4 · Accepted Answer · answered Apr 24 '21 at 13:21

4

All I had to do was create a symlink at /bin/rbash pointing to /bin/bash, and bash must know that is was invoked from a link called rbash and starts itself in restricted mode.

Thanks for your help.

answered Apr 24 '21 at 13:21

Architect

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How can I start a login shell with options?

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