CODEWITHSUNDEEP
bashchroot
Fusion
Fusion

Reputation: 374

chroot + execvp + bash

Update

Got it! See my solution (fifth comment)

Here is my problem:

I have created a small binary called "jail" and in /etc/password I have made it the default shell for a test user.

Here is the -- simplified -- source code:

#define HOME "/home/user"
#define SHELL "/bin/bash"
...
if(chdir(HOME) || chroot(HOME)) return -1;
...
char *shellargv[] = { SHELL, "-login", "-rcfile", "/bin/myscript", 0 };
execvp(SHELL, shellargv);

Well, no matter how hard I try, it seems that, when my test user logs in, /bin/myscript will never be sourced. Similarly, if I drop a .bashrc file in user's home directory, it will be ignored as well.

Why would bash snob these guys?

--

Some precisions, not necessarily relevant, but to clear out some of the points made in the comments:

Upvotes: 8

Views: 1640

Answers (5)

Fusion
Fusion

Reputation: 374

Thanks for your help, guys,

I figured it out:

I forgot to setuid()/setgid(), chroot(), setuid()/setgid() back, then pass a proper environment using execve()

Oh, and, if I pass no argument to bash, it will source ~/.bashrc

If I pass "-l" if will source /etc/profile

Cheers!

Upvotes: 1

Damon Snyder
Damon Snyder

Reputation: 1372

I can identify with wanting to do this yourself, but if you haven't already, check out jail chroot project and jailkit for some drop in tools to create a jail shell.

Upvotes: 2

user149341
user149341

Reputation:

By the time your user is logging in and their shell tries to source this file, it's running under their UID. The chroot() system call is only usable by root -- you'll need to be cleverer than this.

Also, chrooting to a user's home directory will make their shell useless, as (unless they have a lot of stuff in there) they won't have access to any binaries. Useful things like ls, for instance.

Upvotes: 1

ephemient
ephemient

Reputation: 204926

As Jason C says, the exec'ed shell isn't interactive.

His solution will force the shell to be interactive if it accepts -i to mean that (and bash does):

char *shellargv[] = { SHELL, "-i", "-login", ... };
execvp(SHELL, shellargv);

I want to add, though, that traditionally a shell will act as a login shell if ARGV[0] begins with a dash.

char *shellargv[] = {"-"SHELL, "-i", ...};
execvp(SHELL, shellargv);

Usually, though, Bash will autodetect whether it should run interactively or not. Its failure to in your case may be because of missing /dev/* nodes.

Upvotes: 4

dexedrine
dexedrine

Reputation: 169

The shell isn't interactive. Try adding -i to the list of arguments.

Upvotes: 2

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