How can a sourced bash snippet conditionally provide a function to the sourcing shell?

Question

Is it possible to source a Bash snippet, but only actually provide a function from inside it if a certain condition holds true?

So what I am asking is that I can unconditionally source all files from a directory, but the sourced files contain the logic to provide functions to the sourcing shell or not.

Example:

• .bashrc sources whole sub folder .bashrc.d
• .bashrc.d/xyz provides a function adduser2group which works on old systems where usermod doesn't handle -a -G
• .bashrc.d/xyz should provide that function only to the sourcing shell if it's running on such old system.

My current method is to conditionally create an alias named adduser after the Debian program (alias adduser=adduser2group). So I only implement the adduser <user> <group> semantics, but still it's helpful.

Is there a solution that doesn't require this workaround? After all this method means a chance for name clashes which I'd prefer to avoid.

Answer 1

You can use standard shell logic to control whether functions are defined are not.

lib.sh:

if true; then
    foo() {
        echo foo
    }
else
    bar() {
        echo bar
    }
fi

test:

#!/bin/bash

. ./lib.sh
foo
bar

When running it, only foo is defined:

$ bash test
foo
test: line 5: bar: command not found

Substitute if true with more appropriate logic for your application…

Answer 2

You can define the functions you want and, whenever a particular condition holds, you just unset the function:

$ function alpha() { echo $1; }
$ alpha 10
10

Evaluating your condition -- and considering it holds true:

$ if [[ your condition ]]; then unset alpha; fi
$ alpha 10
alpha: command not found