bash alias using sudo returning command not found

Question

I am trying to use the solution of using sudo on my existing aliases as covered in many existing answers already and i am so confused as to why it is not working

alias sudo='sudo '

I keep all my aliases in .bash_aliases. Inside .bash_aliases I have this

function _test {
    echo 'test!'
}

alias test='_test'

I reload the .bashrc each time; source .bashrc but when I run sudo test I always get

sudo: _test: command not found

The only strange thing that happens is that I get the following on reload and not on a new terminal

dircolors: /home/MYHOME/.dircolors: No such file or directory

but i feel this is a red herring.

sudo runs in a clean-ish environment. You can't execute user shell functions like this. The only reason the alias works is because it's expanded before sudo is called. — jordanm
– jordanm, Commented Feb 2, 2021 at 16:54
alias sudo='sudo ' doesn't do anything interesting. Nor does your function alias; you could simply define test() { echo 'test!'; } in the first place, no alias needed. — chepner
– chepner, Commented Feb 2, 2021 at 17:02
@myol chepner's comment does indeed stand. sudo uses execve() to directly invoke the executable that is to be its child process; that's an OS-level syscall. aliases are just an interactive shell construct; they don't work at any other layer, including noninteractive shell scripts (and it's bad form to rely on them at all; while whether they can be exported across a privilege boundary depends on the details of how sudo is configured and it's much better from a security perspective if they're disallowed, exported functions can sometimes work through sudo, whereas aliases never will). — Charles Duffy
– Charles Duffy, Commented Feb 2, 2021 at 17:08
Anyhow -- I'd argue this isn't on-topic here at all; it's a end-user question, not a software-development-specific one. Unix & Linux or Super User are better suited. — Charles Duffy
– Charles Duffy, Commented Feb 2, 2021 at 17:10
@myol You cannot directly sudo a function, an alias or a builtin, only an executable file that is on your disk. The indirect way of doing this is through sudo bash -c .... All this looks like an XY problem to me, though — xhienne
– xhienne, Commented Feb 2, 2021 at 20:14

Charles Duffy · Accepted Answer · 2021-09-17 11:37:03Z

1

As l0b0 says, aliases cannot be used in this way in bash.

However, you can pass a function through (and really, there's basically never a good reason to use an alias instead of sticking to functions alone).

_test() {
    echo 'test!'
}

sudo_test() {
  sudo bash -c "$(declare -f _test)"'; _test "$@"' sudo_test "$@"
}

...will define a command sudo_test that runs the function _test via sudo. (If your real-world version of this function calls other functions or requires access to shell variables, add those other functions to the declare -f command line, and/or add a declare -p inside the same command substitution to generate a textual description of variables your function needs).

answered Sep 17, 2021 at 11:37

Charles Duffy

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l0b0 · Accepted Answer · 2021-02-03 02:22:12Z

0

To run an alias like alias_name it must be exactly the first word in the command, so sudo alias_name will never work. Ditto 'alias_name', \alias_name and other things which eventually expand to the alias name.

answered Feb 3, 2021 at 2:22

l0b0

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