How to read with timeout without making bash consider it an error?

I have a read_user_input.sh script:

#!/bin/bash
# set -e
prompt="bla? [Y/n] "
while true; do read -p "$prompt" -n 1 -s -t 3 reply case $reply in ""|Y|y) echo "bla!!!"; break;; N|n) echo "no bla :( you suck"; break;; *) ;; esac
done

Works as I expected - i.e.

• User inputs "Y" -> bla
• User inputs "y" -> bla
• User hits Enter -> bla
• Timeout -> bla

However, when adding set -e - the read ends with error > 128.

From read --help:

Exit Status:

The return code is zero, unless end-of-file is encountered, read times out (in which case it's greater than 128), a variable assignment error occurs, or an invalid file descriptor is supplied as the argument to -u.

What would be the best way to overcome this?

1. adding || true seems not right, as it would hide any real errors.
2. I also don't want to remove the set -e.
3. One other thing I thought about was handling it in trap but that seems like an overkill

3 Answers

Instead of adding || true, set the $reply to y if there was a timeout:

 read -p "$prompt" -n 1 -s -t 3 reply || { err=$? if (( $err > 128 )) ; then reply=y else exit $err fi }

6

Here is a hack where is used command substitution which forks a subshell where -e is not enabled.

#!/bin/bash
set -e
prompt="bla? [Y/n] "
while true; do case $(read -p "$prompt" -n 1 -s -t 3 reply; echo $reply) in ""|Y|y) echo "bla!!!"; break;; N|n) echo "no bla :( you suck"; break;; *) ;; esac
done

1

Using @choroba answer, I changed my switch case into:

#!/bin/bash
set -e
prompt="bla? [Y/n] "
while true; do read -p "$prompt" -n 1 -s -t 3 reply || (( $? > 128 )) case $reply in ""|Y|y) echo "bla!!!"; break;; N|n) echo "no bla :( you suck"; break;; *) ;; esac
done

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