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?

Solution

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

Comments

[Lennart Rolland]

Gold! But I spent a few minutes noticing the quotation marks :)

[cwingrav]

Inside a function, if you use 'local', it seems to not work. Just FYI.

[Kusalananda]

@cwingrav It works if you separate the local declaration from the assignment into two separate steps: local var; var=$(exit 3); echo "$?"

[cwingrav]

@Kusalananda - agreed. I should have updated my comment to include this (I figured it out shortly after).