"sudo -s <command>" runs command in a shell, but wildcards or metacharacters not working

Question

When using sudo -s (short for the "--shell" option), it is possible to pass "sudo" a command, in which case it will run the command in a shell started by "sudo" as the target user.

(Similarly, sudo -i, also available as the "--login" option, also starts a shell and similarly accepts a command, which behaves the same way.)

Running commands in sudo via a shell can be important in many cases:

• When using wildcards under a directory the current user doesn't have access to, but root does, the command needs to run under a root shell to have the wildcards properly expanded.
• Running a whole pipeline, many commands chained in pipes (|).
• When running a shell built-ins, such as for, if, etc. Running a small whole inline "script" under a single sudo can be useful.

The documentation of the "-s" option says (emphasis mine):

Run the shell specified by the SHELL environment variable if it is set or the shell specified by the invoking user's password database entry. If a command is specified, it is passed to the shell for execution via the shell's -c option. If no command is specified, an interactive shell is executed. Note that most shells behave differently when a command is specified as compared to an interactive session; consult the shell's manual for details.

In other words, when passing sudo -s a command, it is passed to the shell using the -c option, which takes a string with a "script" and then proceeds to execute it as a shell script.

The documentation doesn't really go much further on how to use this option, or to present examples, except to say "consult the shell's manual for details." This implies the command received is passed directly to the shell's -c option. However, as it turns out, that is not the case.

Passing it a shell script with multiple words fails:

$ sudo -s 'ls -ld /var/empty'
/bin/bash: ls -ld /var/empty: No such file or directory

The error message implies it's trying to run the whole string as a simple command... Hmmm, ok, so maybe add spaces would work? Yes that's the case:

$ sudo -s ls -ld /var/empty
drwxr-xr-x. 3 root root 18 Jul 12 21:48 /var/empty

That's not really how the shell's -c works though... Oh well, let's try to use some metacharacters, such as ~, which is a shortcut for the home directory, to see how this behaves. Note the ~ needs to be quoted, to prevent the non-sudo shell from expanding it (in which case it would expand to the home of the non-root user, rather than /root which is expected):

$ sudo -s ls '~'
ls: cannot access '~': No such file or directory

Ok, so that doesn't work, and the error output seems to imply the expansion is not happening, since it's preserving a literal ~ there.

What about wildcards? Not working either:

$ sudo -s ls '/root/*.cfg'
ls: cannot access '/root/*.cfg': No such file or directory

In both these cases, running the command with $SHELL -c works just fine. In this case, $SHELL is bash, so:

$ sudo bash -c 'ls ~'
anaconda-ks.cfg
$ sudo bash -c 'ls /root/*.cfg'
/root/anaconda-ks.cfg

One exception is that variables seem to work on sudo -s, such as:

$ sudo -s echo '$HOME'
/root

So:

• What is going on here?
• Why do wildcards and metacharacters such as ~ do not work on the command passed to sudo -s or sudo -i?
• Given $SHELL -c takes a single string with a script, but sudo -s takes multiple arguments, how is the script assembled from the arguments?
• What is a reliable way to run commands on a shell via sudo?

Answer 1

TL;DR: When taking a command on the "--shell" or "--login" options, sudo will escape most characters (using backslash escapes), including all metacharacters (except for $), also including spaces.

That breaks about every use case where you'd want to use a shell, which makes sudo -s mostly unsuitable to run shell commands that need to run as shell commands.

Instead of sudo -s, use sudo sh -c '...' or sudo bash -c '...' (or whatever the expected $SHELL is.) Instead of sudo -i, use sudo bash -l -c '...' instead (assuming root's shell is bash again.)

---

Looking at the relevant part of the sudo source code, there's this snippet:

/* quote potential meta characters */
if (!isalnum((unsigned char)*src) && *src != '_' && *src != '-' && *src != '$')
    *dst++ = '\\';
*dst++ = *src;

The snippet runs over each character of each argument. If the character is not alphanumeric, underscore, dash or dollar sign, it will be escaped with a backslash.

This means wildcards *, ?, [...], etc. will be escaped to \*, \?, \[...\].

Also metacharacters such as ~, ;, |, etc. will be escaped to \~, \;, \|.

A string with spaces ls /root will be escaped to ls\ /root.

For some reason, $ is an exception in escaping, that's why sudo -s echo '$HOME' works, since the $ will be kept unescaped. (Note this only works for variables, and not even the ${HOME} form, in which case the curly braces will be escaped, which will break the expression.)

---

The consequences of this quoting is that most cases that require running a shell as root can't really benefit from the sudo -s <command> format:

• Wilcards won't be expanded, since they'll be escaped and won't work as wildcards.
• Pipelines won't work, since the pipe | will be escaped and won't work either.
• Commands like for and if typically need to use ;, which will be escaped and won't work either. A newline would be an alternative, but there doesn't seem to be a way to introduce an unescaped newline either.

In short, the form sudo -s <command> only seems to work for cases that don't involve wildcards, pipelines, scriptlets, etc. But then, you typically don't really need a shell to run those commands! Just running sudo <command> without the -s is typically enough for those cases.

It's unclear what the motivation for the sudo implementation. Perhaps to maintain some consistency between handling of commands when adding or removing the -s. Perhaps also because -i precedes -s and in that case there's some small difference in how environment variables are set...

---

WORKAROUND: The workaround is to run the shell's -c explicitly.

This involves knowing what $SHELL is for the target user (which might not match the current user, so using $SHELL directly is not always correct.)

For instance, instead of sudo -s ls -l '/root/*.cfg', use:

$ sudo bash -c 'ls -l /root/*.cfg'

And instead of sudo -i ls -l '~', use:

$ sudo bash -l -c 'ls -l ~'

(bash's -l argument creates a "login" shell, which is equivalent to the one created by sudo -i.)