Script doesn't exit when function output assigned to a variable

Question

A few questions about the sample script below.

I'm calling a function _foo and want to capture the output of it into a variable $bar, but also use the return status (which may not be 0 or 1), or failing that have exit stop the script (when non-zero).

• Why doesn't the exit in function _foo work when called this way? (if ! bar="$(_foo)"). It works when called "normally".
  • the exit will stop the script if I change the if statement to this (but I lose its output): if ! _foo ; then
  • the exit behaves like return and will not stop the script: if ! bar="$(_foo)" ; then
  • Just calling a function without the assignment and an exit will work, however calling it like var="$(func)" doesn't.
• Is there a better way to capture the output of _foo into $bar from the function as well as use return status (for other than 0 or 1, eg a case statement?)

I have a feeling I may need to use trap somehow.

---

Here's a simple example:

#!/usr/bin/env bash

set -e
set -u
set -o pipefail

_foo() {
    local _retval
    echo "baz" && false
    _retval=$?
    exit ${_retval}
}

echo "start"

if ! bar="$(_foo)" ; then
    echo "foo failed"
else
    echo "foo passed"
fi

echo "${bar}"
echo "end"

Here's the ooutput:

$ ./foo.sh 
start
foo failed
baz
end

---

Here's some more examples:

This will exit:

#!/usr/bin/env bash

set -e
set -u
set -o pipefail

func() {
    echo "func"
    exit
}

var=''
func
echo "var is ${var}"
echo "did not exit"

This will not exit:

#!/usr/bin/env bash

set -e
set -u
set -o pipefail

func() {
    echo "func"
    exit
}

var=''
var="$(func)"
echo "var is ${var}"
echo "did not exit"

Answer 1

exit within a function exits the entire script and not just the function (subshells notwithstanding). To expound:

#!/bin/bash
f() {
   exit 3
}
f
exit 0

The above script will terminate with exit code 3, while

#!/bin/bash
f() {
   exit 3
}
(f)
exit 0

will terminate with exit code 0.

The $(command) syntax you are using runs command within a subshell, and exit can only break out as far as the layer that subshell is running within.

If you want to capture the exit code and output of something run within a subshell, that is still available to the environment in which the subshell is initiated:

#!/bin/bash
subshelloutput="$( echo "output"; exit 3 )"
returnval=$?  # captures subshell's exit code
: more stuff follows

Answer 2

You can also do it another way.

_foo () 
{ 
    local _retval
    declare -n output="$1"     # $output is local (by default with declare in a func.)
                               ## and points to $1
    output="baz" && false
    _retval=$?
    echo ${_retval}            # change echo to exit
}

Now call _foo this way:

$ _foo bar
1
$ echo "$bar"
baz

When you change echo to exit in the function definition, the exit will exit the script.