How use local in Bash in conjunction with set -o errexit?

Question

I'm new to Bash and I had a hard time to figure out why when I was using set -o errexit and a command fail, the script was not exiting.

It seems because I declared a local variable!

Please tell me how I can use local variable and set -o errexit at the same time.

Example:

#!/bin/bash
set -o errexit
set -o nounset
set -o pipefail

function test {
 local output=$(ls --badoption)
 echo "error code is $?"
 echo "output=$output"
}

test

Result:

./test.sh
ls: illegal option -- -
usage: ls [-ABCFGHLOPRSTUWabcdefghiklmnopqrstuwx1] [file ...]
error code is 0
output=

But:

#!/bin/bash
set -o errexit
set -o nounset
set -o pipefail

function test {
 output=$(ls --badoption)
 echo "error code is $?"
 echo "output=$output"
}

test

Result:

ls: illegal option -- -
usage: ls [-ABCFGHLOPRSTUWabcdefghiklmnopqrstuwx1] [file ...]

Answer 1

It's because set -o errexit (or running bash with -e) only affects simple commands (as defined in the bash man page). cmd in local output=$(cmd) isn't considered a simple command, so -e doesn't have any affect (and $? doesn't have any relation to how cmd exited). A simple workaround is to split the line by replacing:

local output=$(ls --badoption)

with:

local output

output=$(ls --badoption)

In my testing that does everything you want, it will exit immediately, and if you replace the ls --badoption with just ls, it works, and setting output before calling test then echoing it afterwords shows output really is local to test.