How to disable set -e for an individual command?

Question

The set -e command makes a bash script fail immediately when any command returns an non-zero exit code.

1. Is there an easy and elegant way to disable this behaviour for an individual command within a script?

2. At which places is this functionality documented in the Bash Reference Manual (http://www.gnu.org/software/bash/manual/bashref.html)?

Answer 1

1. Something like this:

   #!/usr/bin/env bash
   
   set -e
   echo hi
   
   # disable exitting on error temporarily
   set +e
   aoeuidhtn
   echo next line
   
   # bring it back
   set -e
   ao
   
   echo next line
   

   Run:

   $ ./test.sh
   hi
   ./test.sh: line 7: aoeuidhtn: command not found
   next line
   ./test.sh: line 11: ao: command not found
   

2. It's described in set builtin help:

   $ type set
   set is a shell builtin
   $ help set
   (...)
   Using + rather than - causes these flags to be turned off.
   

The same is documented here: https://www.gnu.org/software/bash/manual/bashref.html#The-Set-Builtin.

Answer 2

An alternative to unsetting the bail on error would be to force a success no matter what. You can do something like this:

cmd_to_run || true

That will return 0 (true), so the set -e shouldn't be triggered

Answer 3

If you are trying to catch the return/error code (function or fork), this works:

function xyz {
    return 2
}

xyz && RC=$? || RC=$?

Answer 4

If the the "exit immediately shell option" applies or is ignored depends on the context of the executed command (see Bash Reference Manual section on the Set Builtin - thanks to Arkadiusz Drabczyk).

Especially, the option is ignored if a command is part of the test in an if statement. Therefore it is possible to execute a command and check for its success or failure within an "exit immediately context" using an if statement like this:

#!/bin/bash

set -e

# Uncomment next line to see set -e effect:
#blubb

if blubb; then
  echo "Command blubb was succesful."
else 
  echo "Command blubb failed. Exit code: $?"
fi
echo "Script exited normally."

It is possible to omit the "then" statement and use fewer lines:

if blubb; then :;
else echo "Command blubb failed. Exit code: $?"; fi

Answer 5

Another approach, which I find fairly straightforward (and applies to other set options in addition to -e):

Make use of $- to restore settings.

For example:

oldopt=$-
set +e
# now '-e' is definitely disabled.
# do some stuff...

# Restore things back to how they were
set -$oldopt

Though for -e specifically, the options others have mentined (|| true or "put inside an if") may be more idiomatic.

Answer 6

I like to start subshell if want to change something temporarily. Below command demonstrates that first bad_command is ignored and second aborts execution.

bash -c 'set -e ;\
( set +e; bad_command ; echo still here ) ;\
echo even here ; \
bad_command ; \
echo but not here;'

Answer 7

I actually had a similar question recently (though I didn't post, I got around to it), and, from what I can see, it seems like just using set +e before the command and set -e afterward works most elegantly. Here's an example, grabbing the response of the command and not letting the error throw it away.

#!/bin/sh

args=""
for argcol in $*
do
    args="${args} ${argcol}"
done
fortunevar=""
fortfail=""
{
    set +e
    fortunevar=`fortune $args`
    fortfail=$?
    set -e
} &> /dev/null
if [ $fortfail == 0 ]
then
    echo ${fortunevar}
    say ${fortunevar}
else
    echo misfortune: an illegal option was detected!
    echo misfortune: usage: misfortune [-afilosw] [-m pattern][ [#%] file/directory/all]
fi

This grabs the output of 'fortune', checking its exit status, and echoes and says it. I think this is what you were asking for, or at least something similar? Anyway, hope this helps.