Unix & Linux Stack Exchange is a question and answer site for users of Linux, FreeBSD and other Un*x-like operating systems. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to this ref manual:

-E (also -o errtrace)

If set, any trap on ERR is inherited by shell functions, command substitutions, and commands executed in a subshell environment. The ERR trap is normally not inherited in such cases.

However, I must be interpreting it wrongly, because the following does not work:

#!/usr/bin/env bash
# -*- bash -*-

set -e -o pipefail -o errtrace -o functrace

function boom {
  echo "err status: $?"
  exit $?
trap boom ERR

echo $( made up name )
echo "  ! should not be reached ! "

I already know simple assignment, my_var=$(made_up_name), will exit the script with set -e (ie errexit).

Is -E/-o errtrace supposed to work like the above code? Or, most likely, I misread it?

share|improve this question
This is a good question. Replacing echo $( made up name ) with $( made up name ) produces the desired behaviour. I don't have an explanation though. – iruvar Jul 2 '13 at 16:24
I don't know about bash's -E but I do know that -e only effects a shell exit if the error results from the last command in a pipeline. So your var=$( pipe ) and $( pipe ) examples would both represent pipe endpoints whereas pipe > echo would not. My man page says: "1. The failure of any individual command in a multi-command pipeline shall not cause the shell to exit. Only the failure of the pipeline itself shall be considered. " – mikeserv Mar 12 '14 at 10:55
You can make it fail though: echo $(${madeupname?}). But that's set -e. Again, -E is outside my own experience. – mikeserv Mar 12 '14 at 11:03
@mikeserv @1_CR The bash manual @echo indicates that echo always returns 0. This must be factored in the analysis... – On a eu. Mar 15 '14 at 23:28
@illiminE - exactly - it's the last in the pipeline so it won't fail, but the ${var?} if unset will fail unless it is considered to be explicitly tested - according to POSIX - with [ ] || && or similar because it qualifies as a test of its own. If you add ${var?error message} you'll get "error message">&2. – mikeserv Mar 16 '14 at 7:34

Note: zsh will complain about "bad patterns" if you don't configure it to accept "inline comments" for most of the examples here and don't run them through a proxy shell as I have done with sh <<-\CMD.

Ok, so, as I stated in the comments above, I don't know specifically about bash's set -E, but I know that POSIX compatible shells provide a simple means of testing a value if you desire it:

    sh -evx <<-\CMD
    _test() { echo $( ${empty:?error string} ) &&\
        echo "echo still works" 
    _test && echo "_test doesnt fail"
    # END
sh: line 1: empty: error string
+ echo

+ echo 'echo still works'
echo still works
+ echo '_test doesnt fail'
_test doesnt fail

Above you'll see that though I used parameter expansion to test ${empty?} _test() still returns a pass - as is evinced in the last echo This occurs because the failed value kills the $( command substitution ) subshell that contains it, but its parent shell - _test at this time - keeps on trucking. And echo doesn't care - it's plenty happy to serve only a \newline; echo is not a test.

But consider this:

    sh -evx <<-\CMD
    _test() { echo $( ${empty:?error string} ) &&\
            echo "echo still works" ; } 2<<-INIT
            ${empty?function doesnt run}
    _test ||\
            echo "this doesnt even print"
    # END
_test+ sh: line 1: empty: function doesnt run

Because I fed _test()'s input with a pre-evaluated parameter in the INIT here-document now the _test() function doesn't even attempt to run at all. What's more the sh shell apparently gives up the ghost entirely and echo "this doesnt even print" doesn't even print.

Probably that is not what you want.

This happens because the ${var?} style parameter-expansion is designed to quit the shell in the event of a missing parameter, it works like this:


Indicate Error if Null or Unset. If parameter is unset or null, the expansion of word (or a message indicating it is unset if word is omitted) shall be written to standard error and the shell exits with a non-zero exit status. Otherwise, the value of parameter shall be substituted. An interactive shell need not exit.

I won't copy/paste the entire document, but if you want a failure for a set but null value you use the form:

${var :? error message }

With the :colon as above. If you want a null value to succeed, just omit the colon. You can also negate it and fail only for set values, as I'll show in a moment.

Another run of _test():

    sh <<-\CMD
    _test() { echo $( ${empty:?error string} ) &&\
            echo "echo still works" ; } 2<<-INIT
            ${empty?function doesnt run}
    echo "this runs" |\
        ( _test ; echo "this doesnt" ) ||\
            echo "now it prints"
    # END
this runs
sh: line 1: empty: function doesnt run
now it prints

This works with all kinds of quick tests, but above you'll see that _test(), run from the middle of the pipeline fails, and in fact its containing command list subshell fails entirely, as none of the commands within the function run nor the following echo run at all, though it is also shown that it can easily be tested because echo "now it prints" now prints.

The devil is in the details, I guess. In the above case, the shell that exits is not the script's _main | logic | pipeline but the ( subshell in which we ${test?} ) || so a little sandboxing is called for.

And it may not be obvious, but if you wanted to only pass for the opposite case, or only set= values, it's fairly simple as well:

    sh <<-\CMD
    N= #N is NULL
    _test=$N #_test is also NULL and
    v="something you would rather do without"    
    ( #this subshell dies
        echo "v is ${v+set}: and its value is ${v:+not NULL}"
        echo "So this ${_test:-"\$_test:="} will equal ${_test:="$v"}"
        ${_test:+${N:?so you test for it with a little nesting}}
        echo "sure wish we could do some other things"
    ( #this subshell does some other things 
        unset v #to ensure it is definitely unset
        echo "But here v is ${v-unset}: ${v:+you certainly wont see this}"
        echo "So this ${_test:-"\$_test:="} will equal NULL ${_test:="$v"}"
        ${_test:+${N:?is never substituted}}
        echo "so now we can do some other things" 
    #and even though we set _test and unset v in the subshell
    echo "_test is still ${_test:-"NULL"} and ${v:+"v is still $v"}"
    # END
v is set: and its value is not NULL
So this $_test:= will equal something you would rather do without
sh: line 7: N: so you test for it with a little nesting
But here v is unset:
So this $_test:= will equal NULL
so now we can do some other things
_test is still NULL and v is still something you would rather do without

The above example takes advantage of all 4 forms of POSIX parameter substitution and their various :colon null or not null tests. There is more information in the link above, and here it is again.

And I guess we should show our _test function work, too, right? We just declare empty=something as a parameter to our function (or any time beforehand):

    sh <<-\CMD
    _test() { echo $( echo ${empty:?error string} ) &&\
            echo "echo still works" ; } 2<<-INIT
            ${empty?tested as a pass before function runs}
    echo "this runs" >&2 |\
        ( empty=not_empty _test ; echo "yay! I print now!" ) ||\
            echo "suspiciously quiet"
    # END
this runs
echo still works
yay! I print now!

It should be noted that this evaluation stands alone - it requires no additional test to fail. A couple more examples:

    sh <<-\CMD
    ${empty?null, no colon, no failure}
    unset empty
    echo "${empty?this is stderr} this is not"
    # END
sh: line 3: empty: this is stderr

    sh <<-\CMD
    _input_fn() { set -- "$@" #redundant
            echo ${*?WHERES MY DATA?}
            #echo is not necessary though
            shift #sure hope we have more than $1 parameter
            : ${*?WHERES MY DATA?} #: do nothing, gracefully
    _input_fn heres some stuff
    _input_fn one #here
    # shell dies - third try doesnt run
    _input_fn you there?
    # END
heres some stuff
sh: line :5 *: WHERES MY DATA?

And so finally we come back to the original question : how to handle errors in a $(command substitution) subshell? The truth is - there are two ways, but neither is direct. The core of the problem is the shell's evaluation process - shell expansions (including $(command substitution)) happen earlier in the shell's evaluation process than does current shell command execution - which is when your errors could be caught and trapped.

The issue the op experiences is that by the time the current shell evaluates for errors, the $(command substitution) subshell has already been substituted away - no errors remain.

So what are the two ways? Either you do it explicitly within the $(command substitution) subshell with tests as you would without it, or you absorb its results into a current shell variable and test its value.

Method 1:

    echo "$(madeup && echo \: || echo '${fail:?die}')" |\
          . /dev/stdin

sh: command not found: madeup
/dev/stdin:1: fail: die

    echo $?


Method 2:

    var="$(madeup)" ; echo "${var:?die} still not stderr"

sh: command not found: madeup
sh: var: die

    echo $?


This will fail regardless of the number of variables declared per line:

   v1="$(madeup)" v2="$(ls)" ; echo "${v1:?}" "${v2:?}"

sh: command not found: madeup
sh: v1: parameter not set

And our return value remains constant:

    echo $?


    trap 'printf %s\\n trap resurrects shell!' ERR
    v1="$(madeup)" v2="$(printf %s\\n shown after trap)"
    echo "${v1:?#1 - still stderr}" "${v2:?invisible}"

sh: command not found: madeup
sh: v1: #1 - still stderr

    echo $?
share|improve this answer
For instance, practically his exact specification: echo ${v=$( madeupname )} ; echo "${v:?trap1st} ! is not reached !" – mikeserv Mar 27 '14 at 6:38
To summarize: These parameter expansions are a great way to control if variables are set etc. In respect of the exit status of echo, I noticed that echo "abc" "${v1:?}" doesn't seem to execute(abc never printed). And the shell returns 1. This is true with or without a command even("${v1:?}" directly on the cli). But for OP's script, all that was required to trigger the trap is placing his variable assignment containing a substitution for a nonexistent command alone on a single line. Otherwise the behavior of echo is to return 0 always, unless interrupted like with the tests you explained. – On a eu. Mar 27 '14 at 22:45
Just v=$( madeup ). I don't see what is unsafe with that. It's just an assignment, the person mispelled the command for instance v="$(lss)". It errors. Yes you can verify with the error status of the last command $? - because it is the command on the line (an assignment with no command name) and nothing else - not an argument to echo. Plus, here it's trapped by the function as !=0, so you get feedback twice. Otherwise surely as you explain there is a better way to do this orderly within a framework but OP had 1 single line: an echo plus his failed substitution. He was wondering about echo. – On a eu. Mar 28 '14 at 2:17
Yes, simple assignment is mentioned in the question as well as a for granted, and while i couldnt offer specific advice on how to make use of -E, i did try to show similar results to the demonstrated issue were more than possible without resorting to bashisms. In any case, it is specifically the issues you mention - like single assignment per line - that make such solutions difficult to handle in a pipeline, which i also demonstrated how to handle. Though it's true what you say - there is nothing unsafe about simply assigning it. – mikeserv Mar 28 '14 at 2:29
Good point - perhaps more effort is called for. Ill think on it. – mikeserv Mar 28 '14 at 3:11

If set, any trap on ERR is inherited by shell functions, command substitutions, and commands executed in a subshell environment

In your script its a command execution(echo $( made up name )). In bash commands are delimited with either ; or with new line. In the command

echo $( made up name )

$( made up name ) is considered as a part of the command. Even if this part fails and return with error, the whole command executes successfully as echo don't know about it. As the command return with 0 no trap triggered.

You need to put it in two commands, assignment and echo

echo $var
share|improve this answer
echo $var doesn't fail - $var is expanded to empty before ever echo actually gets a look at it. – mikeserv Mar 17 '14 at 1:47

Please consider the difference in the output of the following sets of commands which leverage the ? parameter, expanding to the exit status of the most recently executed foreground pipeline:

# grep . $(made) $?; echo $? ${PIPESTATUS[@]}
bash: made: command not found
grep: 127: No such file or directory
2 2

# echo $(made) $?; echo $? ${PIPESTATUS[@]}
bash: made: command not found
0 0

The echo bash builtin always returns 0:

echo [-neE] [arg …]

Output the args, separated by spaces, terminated with a newline. The return status is always 0. (...)

Command substitution happens in a subshell environment and furthermore, if a command is not found, the child process created to execute it returns a status of 127. And so it did in our examples. The grep command exits with a status of 2(an error occurred that is not related to not finding lines). The behavior of the echo command is predictable as it's documented in the bash manual: it returns 0 whatever happens to the expansion of its arguments. So there is no error to trap in your example.

The simple command: a reminder

You mentioned that variable assignment containing a command substitution to a command that doens't exist will trigger exiting of the shell with set -e. That is not the case with optional variable assignment preceding a command name in a simple command:

# var=$(made) echo $?; echo $? ${PIPESTATUS[@]}
bash: made: command not found
0 0

But is indeed the case with a simple command which doesn't contain a command name i.e. only variable assignment(s):

var=$(made); echo $? ${PIPESTATUS[@]}
bash: made: command not found
127 127

but not if there is more that one command substitution in those assignments and the last one succeeds:

var=$(made) var2=$(ls); echo $? ${PIPESTATUS[@]}
bash: made: command not found
0 0

Again, this behavior is entirely documented:

If there is a command name left after expansion, execution proceeds as described below. Otherwise, the command exits. If one of the expansions contained a command substitution, the exit status of the command is the exit status of the last command substitution performed. If there were no command substitutions, the command exits with a status of zero.

In our last example, the two variable assignments constitute a (simple) command obviously, but not one with a command name after the assignment.


Keeping one variable assignment per line should allow for any command substitution it contains to be trapped in case of an error. The echo builtin will return 0 as its exit status, and the exit status of its arguments can never taint that. The shell can fail for other reasons, for instance in the case of a specific method designed to validate if variables are set (as demonstrated by another answer to this Q) which would be evaluated prior in the order of execution. But the fact remains: the behavior of the shell when a command is not found, whether or not inside a command substitution, is predictable and documented - the child process returns 127 but echo proceeds. This is why the substitution has to be tested either in a single assignment or through some other means. This understanding is a prerequisite to exploring what difference the set options will make in relation to the trap in your script.

share|improve this answer
I made echo error. I also made null (or true) error. – mikeserv Mar 26 '14 at 13:56
Echo is not a separate process - echo is a shell builtin. ${v?} is the same process... and printf... and trap... All builtins. All the same process. So it's impossible to use those kinds of superlatives when the same process incorporates a test that will always kill itself upon failure, unless you consider trap, which will always restrict the process from killing itself and so on and so on... – mikeserv Mar 27 '14 at 0:01
To summarize: "echo returns 0 if it returns at all". It doesn't get to return anything if one of its arguments is something like "${v1:?}" and v1 is not set. The shell returns 1 and exits. In general, a command returns a status of success/failure i.e. 0 or 1. When an argument composed of a command substitution for a command that doesn't exist is echoed, echo will never fail, but the argument will return 127. Another command than echo, like grep . "$(badcommand)" will return 2 (without quotes this expands to nothing and goes to stdin, where EOF would trigger the trap in this case). Thank you! – On a eu. Mar 27 '14 at 21:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.