Bash scripting: How to ensure script exits on any error in function called with an "or" (||) condition?

Question

Here is an example script:

#!/bin/bash
set -euo pipefail

function_test () {
        echo "Everything still ok."
        thisdoesnotexist
        echo "It continues running..."
}

function_test || { echo "Error occured in function_test."; exit 1; }

I expect this script to exit on "thisdoesnotexist", since set -e is activated. Either the error message "Error occured in function_test." should appear, or at least the script should be exited. What happens instead, is that the script keeps running:

$ ./testscript.sh 
Everything still ok.
./testscript.sh: Zeile 6: thisdoesnotexist: Kommando nicht gefunden.
It continues running...

I figure this is because I'm using the function in an "or" (||) context, according to the manual for set:

Exit immediately [...] unless the command that fails is part of an until or while loop, part of an if statement, part of a && or || list, [...]

What would be the best way to handle this in a script? I have removed the "||" and used an error trap like this:

#!/bin/bash
set -Eeuo pipefail

errorhandler () {
        echo "Errorhandler called. Exiting."
        exit 1
}

trap "errorhandler" ERR

function_test () {
        echo "Everything still ok."
        thisdoesnotexist
        echo "It continues running..."
}

function_test

Which does work:

$ ./testscript.sh 
Everything still ok.
./testscript.sh: Zeile 13: thisdoesnotexist: Kommando nicht gefunden.
Errorhandler called. Exiting.

In this case though it seems impossible to output a user friendly message (which, for example, contains the step in the script, where the error happened). For this I would at least have to pass on the function name to the error trap. I tried that but found no working solution.

Do you have a better suggestion?

tDarkCrystal · Accepted Answer · 2019-08-11 01:25:41Z

2

function_test || { echo "Error occured in function_test."; exit 1; }

The OR never happens because of the functions return status. The last echo statement from the function causes a return status of 0, since it successfully ran.

If you add a non-zero return on the command that isn't successful, then it will cause the OR statement to run properly.

function_test () {
        echo "Everything still ok."
        thisdoesnotexist || return 1;
        echo "It continues running..."
}

edited Aug 11, 2019 at 1:25

answered Aug 11, 2019 at 1:19

tDarkCrystal

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Comments

KamilCuk · Accepted Answer · 2019-08-11 10:53:47Z

0

Not very expendable, but preferable in some cases. Export flags and your function, and run it in a subshell. Handle bash error value.

#!/bin/bash
set -euo pipefail

function_test () {
    echo "Everything still ok."
    thisdoesnotexist
    echo "It continues running..."
}

export -f function_test
export SHELLOPTS
bash -c function_test || { echo "Error occured in function_test."; exit 1; }

answered Aug 11, 2019 at 10:53

KamilCuk

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Comments

Léa Gris · Accepted Answer · 2019-08-11 11:11:02Z

0

If you want to do stuffs in case of error and you know something may go wrong then:

function_test () {
        echo "Everything still ok."
        thisdoesnotexist || {
          # handle thisdoesnotexist failure
          Echo "thisdoesnotexist failed" >&2
          exit 1
        }
        echo "It continues running..."
        if ! thisdoesnotexisteither; then
          # Handle fail for thisdoesnotexisteither
          Echo "thisdoesnotexisteither failed" >&2
          exit 1
        fi
}

Now for the purpose of testing error handling in Bash, you can use this intelligent erroring function:

#!/usr/bin/env bash

# Generate an error
# @Params
# $1: Numeric error return code
# ${2[@]}: Messages
# @Output
# &1: Generic success message
# &2: Generic or passed error messages
# @Return
# $?: Intended return code
erroring() {
  local rc=1
  if [ "$#" -gt 0 ]; then
    rc="$1"
    shift
    if printf '%d' "$rc" >/dev/null 2>&1; then
      # rc is an integer
      if [ "$rc" -gt 0 ]; then
        printf $"Intended failure with code: %d\\n" "$rc" >&2
        if [ "$#" -gt 0 ]; then
          # There are supplied messages to print"
          printf $"%d messages:\\n" "$#" >&2
          printf '%q\n' "$@" >&2
        fi
      else
        echo $"Intended success"
      fi
    else
      printf $"Invalid non-numeric return code parameter: %q\\n" "$rc" >&2
      rc=7
    fi
  else
    erroring "$rc"
  fi
  return "$rc"
}

edited Aug 11, 2019 at 11:11

answered Aug 11, 2019 at 0:29

Léa Gris

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2 Comments

KamilCuk Over a year ago

the last printf $'Invalid - this is intended to be printf $"Invalid ? What is the point of calling erroring recursively? Wouldn't just if [ "$#" -gt 0 ]; then rc=$1; else rc=1; fi be simpler?

Léa Gris Over a year ago

This is sugar for the erroring function to fail on its own way because it was passed a non-numeric return code. Like calling erroring "Error message" rather than calling erroring 1 "Error message"

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