Krdytkyu

I have this in a bash script:

exit 3;



exit_code="$?"



if [[ "$exit_code" != "0" ]]; then

    echo -e "${r2g_magenta}Your r2g process is exiting with code $exit_code.${r2g_no_color}";

    exit "$exit_code";

fi

It looks like it will exit right after the exit command, which makes sense.
I was wondering is there some simple command that can provide an exit code without exiting right away?

I was going to guess:

exec exit 3

but it gives an error message: exec: exit: not found. 
What can I do? :)

If you have a script that runs some program
and looks at the program's exit status (with $?),
and you want to test that script by doing something
that causes $? to be set to some known value (e.g., 3), just do

(exit 3)

The parentheses create a sub-shell. 
Then the exit command causes that sub-shell
to exit with the specified exit status.

exit is a bash built-in, so you can't exec it. Per bash's manual:

Bash's exit status is the exit status of the last command executed in the script. If no commands are executed, the exit status is 0.

Putting all this together, I'd say your only option is to store your desired exit status in a variable and then exit $MY_EXIT_STATUS when appropriate.

You can write a function that returns the status given as argument, or 255 if none given. (I call it ret as it "returns" its value.)

ret() { return "${1:-255}"; }

and use ret in place of your call to exit. This is avoids the inefficiency of creating the sub-shell in the currently accepted answer.

About exec exit 3... it would try to run an external command called exit, but there isn't one, so you get the error. It has to be an external command instead of one built in to the shell, since exec replaces the shell completely. Which also means that even if you had an external command called exit, exec exit 3 would not return to continue your shell script, since the shell wouldn't be there any more.

You can do this with Awk:

awk 'BEGIN{exit 9}'

Or Sed:

sed Q9 /proc/stat