Calling function from another function in Bash

Question

I have the following Bash script:

func_1() {
  local arg="$1"
  local var="$(curl $someUrl "$arg")"
  echo "$var"
}

main() {
  # ...
  # some_arg defined
  output="$(func_1 "$some_arg")"

  echo "$output"
}

main

However, after running this script, instead of getting the result of func_1 being assigned to output and then printed, the func_1 is being executed when echoed.

What do I have to modify to execute the function, assign the result and then use the variable in later part of the script?

Change output="$(func_1 "$some_arg")" to output=$(func_1 "$some_arg"). Your quotes are all messed up. They do not nest. — Jack
– Jack, Commented Jul 13, 2021 at 13:53
@Jack, they aren't nested - inside those parens is a whole different subshell. The syntax is valid. Try echo "$( echo "date;date" )" vs echo "$( echo date;date )". — Paul Hodges
– Paul Hodges, Commented Jul 13, 2021 at 13:59
Try output="$(func_1 "$some_arg" 2>&1)". What output is beeing outputted from func_1? — KamilCuk
– KamilCuk, Commented Jul 13, 2021 at 14:02
@Forin So how do you know that func_1 is not being executed? I put in some echo directing output to /dev/stderr, and it all executed as written. — Jack
– Jack, Commented Jul 13, 2021 at 15:01

Paul Hodges · Accepted Answer · 2021-07-13 14:05:03Z

1

Clarify your testing with some clear indicators.

func_1() {
  local arg="$1"
  local var="$(echo "quoted '$1'")"
  echo "$var"
}

main() {
  local some_arg='this literal'
  output="$(func_1 "$some_arg")"

  echo "[$output]"
}

Now if I run it:

$: set -x; main; set +x;
+ main
+ local 'some_arg=this literal'
++ func_1 'this literal'
++ local 'arg=this literal'
+++ echo 'quoted '\''this literal'\'''
++ local 'var=quoted '\''this literal'\'''
++ echo 'quoted '\''this literal'\'''
+ output='quoted '\''this literal'\'''
+ echo '[quoted '\''this literal'\'']'
[quoted 'this literal']
+ set +x

Seems to be working fine.

answered Jul 13, 2021 at 14:05

Paul Hodges

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1 Comment

Jack Over a year ago

Worked for me, too. I don't see a problem.

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