How to return an array in bash without using globals?

Question

I have a function that creates an array and I want to return the array to the caller:

create_array() {
  local my_list=(\"a\", \"b\", \"c\")
  echo \"${my_li         


        

Answer 1

Use the technique developed by Matt McClure: http://notes-matthewlmcclure.blogspot.com/2009/12/return-array-from-bash-function-v-2.html

Avoiding global variables means you can use the function in a pipe. Here is an example:

#!/bin/bash

makeJunk()
{
   echo 'this is junk'
   echo '#more junk and "b@d" characters!'
   echo '!#$^%^&(*)_^&% ^$#@:"<>?/.,\\"'"'"
}

processJunk()
{
    local -a arr=()    
    # read each input and add it to arr
    while read -r line
    do 
       arr+=('"'"$line"'" is junk') 
    done;

    # output the array as a string in the "declare" representation
    declare -p arr | sed -e 's/^declare -a [^=]*=//'
}

# processJunk returns the array in a flattened string ready for "declare"
# Note that because of the pipe processJunk cannot return anything using
# a global variable
returned_string="$(makeJunk | processJunk)"

# convert the returned string to an array named returned_array
# declare correctly manages spaces and bad characters
eval "declare -a returned_array=${returned_string}"

for junk in "${returned_array[@]}"
do
   echo "$junk"
done

Output is:

"this is junk" is junk
"#more junk and "b@d" characters!" is junk
"!#$^%^&(*)_^&% ^$#@:"<>?/.,\\"'" is junk