How does bash know how it is being invoked?

Question

I've installed jailkit on Ubuntu 12.04 and I have set up a user's shell to /bin/bash - but when it is invoked it runs /etc/bash.bashrc instead of /etc/profile

If you haven't used jailkit before here's the gist of it:

1. A "jailed" version of the system root is created somewhere, like /home/jail
2. Jailed users home directories are moved inside that folder like /home/jail/home/testuser
3. Relavant configuration files are copied to /home/jail/etc/ - including a limited /etc/passwd
4. Programs that you want to allow access to are copied to the corresponding directories, like /bin/bash
5. When a jailed user logs in they are chrooted to /etc/jail/ and can't see any files above that

So I have a testuser who has an entry in /etc/passwd like this:

testuser:x:1002:1003::/home/jail/./home/testuser:/usr/sbin/jk_chrootsh

In the file /home/jail/etc/passwd there is an entry like:

testuser:1001:1003::/home/testuser:/bin/bash

I've read though the bash(1) and so I think the problem is that bash thinks it is not being invoked as a login shell:

When bash is invoked as an interactive login shell, or as a non-interactive shell with the --login option, it first reads and executes commands from the file /etc/profile, if that file exists.

I get that bash is actually being invoked by /usr/sbin/jk_chrootsh but I don't understand how bash is determining what type of shell it is, and what set of startup files it should run.

I'd like to see if I can troubleshoot this - but I don't understand:

How does bash know how it is being invoked?

ps: I also looked into login(1) without much luck.

Answer 1

Normally bash knows that it's a login shell because when the login program invokes it, it tells bash that its name is -bash. That name is in argv[0], the zeroth command line argument, which is conventionally the way the user invoked the program. The initial hyphen is a convention to tell a shell that it's a login shell. Bash also behaves as a login shell if you pass it the option --login or -l. See Difference between Login Shell and Non-Login Shell? for more details.

As of Jailkit 2.16, jk_chrootsh reads the absolute path to the shell to invoke from various sources, and passes this path as argv[0], and passes its own command line arguments down to that shell. In the normal use case where jk_chrootsh is itself used in /etc/passwd, there is no way to pass an argument such as -l. Since the absolute path doesn't begin with -, there is no way to make jk_chrootsh invoke a login shell, short of using a tiny intermediate program.

#include <unistd.h>
int main () {
    execl("/bin/bash", "-bash", NULL);
    return 127;
}

I would have expected jk_chrootsh to have an easy way of invoking a login shell. I suggest making a feature request.

Answer 2

login calls the login command/shell of the user with its argv[0] starting with a -. Shells check their argv[0] to determine if they're being called as a login shell.

As @slm says, it's clearly specified in the "Invocation" section of the bash manual.

In addition, a few shells like csh, tcsh, ksh, zsh, yash, bash and some variants of the Almquist shell support the -l option to enable the login mode without having to mingle with the first argument. That is not used by login, but you can use it if you want to simulate a login shell from something (like most shells) where it's difficult to run a command with an arbitrary argv[0]. I've seen it used by graphical login managers.

Answer 3

Take a look at the bash man page. They discuss the differences in how it can be invoked there. The section is called INVOCATION. The 2 primary ways it get's invoked are as a login shell (bash -l) and as a interactive shell (bash -i).

Take a look at this other Unix and Linux Q&A titled: Difference between Login Shell and Non-Login Shell?. It pretty much covers exactly what you're asking about.