Capture exit code of exit command

Question

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? :)

Solution 1

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.

Solution 2

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.

Solution 3

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.

Some measurements.

time bash -c 'for i in {1..10000} ; do (exit 3) ; done ; echo $?'

on my machine takes about 3.5 seconds.

 time bash -c 'ret(){ return $1 ; } ; for i in {1..10000} ; do ret 3 ; done ; echo $?'

on my machine takes about 0.051 seconds, 70 times faster. Putting in the default handling still leaves it 60 times faster. Obviously the loop has some overhead. If I change the body of the loop to just be : or true then the time is halved to 0.025, a completely empty loop is invalid syntax. Adding ;: to the loop shows that this minimal command takes 0.007 seconds, so the loop overhead is about 0.018. Subtracting this overhead from the two tests shows that the ret solution is over 100 times faster.

Obviously this is a synthetic measurement, but things add up. If you make everything 100 times slower than they need to be then you end up with slow systems. 0.0

Solution 4

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.

Solution 5

You can do this with Awk:

awk 'BEGIN{exit 9}'

Or Sed:

sed Q9 /proc/stat

View more solutions

Comments

[Admin]

hmmm what do you think about @G-man's idea?

[solarshado]

Maybe I misunderstood what you're trying to accomplish. If you're just trying to set $? (though I'm not really sure why you would), that does seem like a solid answer. If you just want to set it to some unsuccessful value, false is another option.

[icarus]

@iBug The extra space is not needed.

[Jerb]

I guess you could do exec bash -c "exit 3", but at the moment I can't think of any reason to do that as opposed to just exit 3.

[ilkkachu]

...or with a shell: sh -c 'exit 9'

[user1730706]

@ilkkachu if youre going to do that you might as well do the (exit 9) in the accepted answer

[G-Man Says 'Reinstate Monica']

Good point about the inefficiency of creating the sub-shell.  I've read that some shells might be smart enough to optimize the fork out in cases like this, but that bash isn't one of them.

[Centimane]

Also, for debugging purposes it'd be just as simple to set exit_code="3" for testing

[G-Man Says 'Reinstate Monica']

Yes, wjandrea pointed that out yesterday.