Can I get the exit code from a sub shell launched with $(command)?

Question

I am setting a variable like this:

myvar=$(command --params)

I want to check the exit code ($?) of my command afterwards. Checking $? like this always returns 0 because it successfully set the variable to the output of the command.

Is it possible to get the return value of command?

Related: http://unix.stackexchange.com/questions/340997/assignments-are-like-commands-with-an-exit-status-except-when-theres-command-su — Kusalananda, Feb 22 '17 at 17:30
The premise here is wrong; $? only returns 0 if the command succeeded. It's not simply "always 0 because of successful variable assignment." — Wildcard, Feb 22 '17 at 17:47

score 46 · Accepted Answer · answered Feb 22 '17 at 17:27

46

Yes, it is possible without even getting too far out of your way:

$ $(exit 3); echo $?
3

$ foo="$(echo bar; exit 3)"; echo $?; echo $foo
3
bar

answered Feb 22 '17 at 17:27

DopeGhoti

76,081

Gold! But I spent a few minutes noticing the quotation marks :) – Mr. Developerdude Apr 17 '18 at 08:03
9

Inside a function, if you use 'local', it seems to not work. Just FYI. – cwingrav Jul 09 '19 at 10:56
16

@cwingrav It works if you separate the local declaration from the assignment into two separate steps: local var; var=$(exit 3); echo "$?" – Kusalananda Dec 28 '19 at 07:24
@Kusalananda - agreed. I should have updated my comment to include this (I figured it out shortly after). – cwingrav Dec 29 '19 at 12:18

Can I get the exit code from a sub shell launched with $(command)?

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