[SOLVED] Bash error trap calling erroneous helper function

Issue

I want to add an error trap function to my bash script. I cannot figure out how to call helper functions (which may themselves contain errors) from the error trap function without very weird behavior.

Running the following code snippet

#!/bin/bash

# enable bash strict mode
# -u: treat unset variables as an error
# -o pipefail: set the exit status of a pipeline to the exit status of the rightmost non-zero exit status
# -e: Script exits immediately on errors
set -e -u -o pipefail

on_error() {
    echo "Start error"
    local helper_exit_status=0
    helper || helper_exit_status=$?
    echo "Finish error $helper_exit_status"
}

helper() {
    echo "Start helper"
    ls /nonexistent-directory-2
    echo "Finish helper"  # <-- I would expect this to not be executed due to "set -e"
}

trap 'on_error' ERR

echo "Start main"

ls /nonexistent-directory-1

echo "Finish main"

with bash yields

$ bash ./test.sh
Start main
ls: cannot access '/nonexistent-directory-1': No such file or directory
Start error
Start helper
ls: cannot access '/nonexistent-directory-2': No such file or directory
Finish helper
Finish error 0

I think what is happening is this:

1. Print "Start main"
2. ls /nonexistent-directory-1 has non-zero exit status
3. This triggers a call to on_error
4. Print "Start error"
5. Call helper, where the "||" means that subsequent non-zero exit status is ignored
6. Print "Start helper"
7. ls /nonexistent-directory-1 has non-zero exit status (ignored)
8. Print "Finish helper"
9. The last command was echo, which had non-zero exit status, so helper_exit_status remains 0

How can I fix this such that:

• My script still calls on_error when encountering errors
• on_error can call helper functions like helper
• An error in helper immediately aborts helper (this is the expected behavior due to set -e, and I should not need to modify helper to achieve this as it can also be called from other locations)
• After helper encounters an error, on_error continues its execution but learns about the error due to a non-zero helper_exit_status

---

My bash version:

$ bash --version
GNU bash, version 5.0.17(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2019 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.

Solution

set -e and Functions: When set -e is active, Bash will exit immediately if a command exits with a non-zero status, unless the non-zero status is part of an if statement, part of a test within a list construct (&& or ||), or the command's return status is being inverted with !.

Helper Function: When the helper function is called with ||, the non-zero status of ls is considered to be controlled by the ||. Therefore, the script doesn't exit, and the echo "Finish helper" command is executed.

Error Status in on_error Function: The command helper || helper_exit_status=$? sets helper_exit_status to 0 because the last command (echo) in the helper function exits successfully.

Here's an updated version of your script:

    #!/bin/bash

set -e -u -o pipefail

on_error() {
    echo "Start error"
    local helper_exit_status=0
    helper
    helper_exit_status=$?
    echo "Finish error $helper_exit_status"
}

helper() {
    echo "Start helper"
    ls /nonexistent-directory-2 || return $?
    echo "Finish helper"
}

trap 'on_error' ERR

echo "Start main"

ls /nonexistent-directory-1

echo "Finish main"

Answered By - Przemysław Doczkal

Answer Checked By - Senaida (WPSolving Volunteer)