Shell Script – How to Exit Shell Script from a Subshell

exitshellshell-scriptsubshell

Consider this snippet:

stop () {
    echo "${1}" 1>&2
    exit 1
}

func () {
    if false; then
        echo "foo"
    else
        stop "something went wrong"
    fi
}

Normally when func is called it will cause the script to terminate, which is the intended behaviour. However, if it's executed in a sub-shell, such as in

result=`func`

it will not exit the script. This means the calling code has to check the exit status of the function every time. Is there a way to avoid this? Is this what set -e is for?

Best Answer

You could kill the original shell (kill $$) before calling exit, and that'd probably work. But:

it seems rather ugly to me
it will break if you have a second subshell in there, i.e., use a subshell inside a subshell.

Instead, you could use one of the several ways to pass back a value in the Bash FAQ. Most of them aren't so great, unfortunately. You may just be stuck checking for errors after each function call (-e has a lot of problems). Either that, or switch to Perl.

Related Solutions

Bash – Last exit status from within script without source

I'm not quite sure where that will lead to, but you can do:

alias dummy='sh dummy $?'
the_program_with_errors
dummy

and your dummy script would contain:

echo $1

An approach without alias is to use a shell function:

function dummy { sh dummy $? ;}

With that definition you can get the following behaviour (simulated with true, false, and a subshell process):

$ true 
$ dummy
0
$ false
$ dummy
1
$ (exit 42)
$ dummy    
42

Bash – Issue with Boolean Tests && and ||

Because the last command executed (the echo) succeeded. If you want to group the commands, it's clearer to use an if statement:

if ! step 1 ; then
   echo >&2 "Failed to execute step 1"
   exit 2
fi

You could also group the error message and exit with { ... } but that's somewhat harder to read. However that does preserve the exact value of the exit status.

step 1 || { echo >&2 "step 1 failed with code: $?"; exit 2; }

Note, I changed the && to a semicolon, since I assume you want to exit even if the error message fails (and output those errors on stderr for good practice).

For the if variant to preserve the exit status, you'd need to add your code to the else part:

if step 1; then
  : OK, do nothing
else
  echo >&2 "step 1 failed with code: $?"
  exit 2
fi

(that also makes it compatible with the Bourne shell that didn't have the ! keyword).

As or why the commands group like the do, the standard says:

An AND-OR list is a sequence of one or more pipelines separated by the operators "&&" and "||".

The operators "&&" and "||" shall have equal precedence and shall be evaluated with left associativity.

Which means that something like somecmd || echo && exit acts as if somecmd and echo were grouped together first, i.e. { somecmd || echo; } && exit and not somecmd || { echo && exit; } .

Best Answer

Related Solutions

Bash – Last exit status from within script without source

Bash – Issue with Boolean Tests && and ||

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