bash function with return value AND return string

Question

First my sourcecode to test the returning method of bash functions:

function oidev.fnc.bash.example.fncreturnstringvalue
{
  echo "ABC"
  return 1
}

and the caller function:

function oidev.fnc.bash.example.fncreturnstringvaluecaller
{
  local FNCVALUE
  local FNCSTRING
  FNCSTRING=$(oidev.fnc.bash.example.fncreturnstringvalue && FNCVALUE="$?" || FNCVALUE="$?")
   # String OK but Value empty
  echo "VALUE : $FNCVALUE" 
  echo "STRING: $FNCSTRING"
  unset FNCVALUE
  unset FNCSTRING
  oidev.fnc.bash.example.fncreturnstringvalue && FNCVALUE="$?" || FNCVALUE="$?"
    # Value OK and String on STDOUT
  echo "VALUE : $FNCVALUE" 
  echo "STRING: $FNCSTRING"
}

The output in bash:
VALUE :
STRING: ABC
ABC
VALUE : 1
STRING: (surely empty, but echoed from the function) `

And now my simple question:
Is it possible to get the returning string and returning value into two different variables by a single line construction?

I don't want to use globals, nore the use of if $? after the sub-call!
Many thanks for help, and sorry for my german-english!

Answer 1

Try this, maybe helps:

func() {
    echo 'abc'
    return 111
}

read var1 var2 <<<$(func; echo $?)
echo "var1 $var1"
echo "var2 $var2"

will print:

var1 abc
var2 111

workaround for the "spaces problem" - now, haven't better idea as use an helper function like this

func() {
    echo 'abc def'
    return 111
}

runner() {
    ret=$(func)
    echo $? $ret
}

read -r var2 var1 <<<$(runner func)
echo "var1 $var1"
echo "var2 $var2"

or using a helper variable and switching the variable order

func() {
    echo 'abc def'
    return 111
}
read -r var2 var1 <<<$(res=$(func); echo $? $res)
echo "var1 $var1"
echo "var2 $var2"

both will return

var1 abc def
var2 111