Returning array from a Bash function

This won't work as expected when there are whitespaces in the arrays:

function create_some_array() {
    local -a a=()
    for i in $(seq $1 $2); do
        a[i]="$i $[$i*$i]"
    done
    echo ${a[@]}
}

and worse: if you try to get array indices from the outside "a", it turns out to be a scalar:

echo ${!a[@]}

even assignment as an array wont help, as possible quoting is naturally removed by the echo line and evaluation order cannot be manipulated to escape quoting: try

function create_some_array() {
...
    echo "${a[@]}"
}

a=($(create_some_array 0 10))
echo ${!a[@]}

Still, printf seems not to help either:

function create_some_array() {
...
    printf " \"%s\"" "${a[@]}"
}

seems to produce correct output on one hand:

$ create_some_array 0 3; echo
 "0 0" "1 1" "2 4" "3 9"

but assignment doesn't work on the other:

$ b=($(create_some_array 0 3))
$ echo ${!b[@]}
0 1 2 3 4 5 6 7

So my last trick was to do assignment as follows:

$ eval b=("$(create_some_array 0 3)")
$ echo -e "${!b[@]}\n${b[3]}"
0 1 2 3
3 9

Tataaa!

P.S.: printf "%q " "${a[@]}" also works fine...

This works fine as described. The most likely reason it doesn't work in your actual code is because you happen to run it in a subshell:

cat textfile | create_some_array
echo ${a[*]}

would not work, because each element in a pipeline runs in a subshell, and

myvalue=$(create_some_array)
echo ${a[*]}

would not work, since command expansion happens in a subshell.

You can make an array local to a function, and then return it:

function create_some_array(){
    local -a a=()
    for i in $(seq $1 $2); do
        a[i]=$i
    done
    echo ${a[@]}
}

declare -a a=()

a=$(create_some_array 0 10)

for i in ${a[@]}; do
   echo "i = " $i
done

Hi here is my solution:

show(){
    local array=()
    array+=("hi")
    array+=("everything")
    array+=("well?")
    echo "${array[@]}"
}

for e in $(show);do
    echo $e
done

Try this code on: https://www.tutorialspoint.com/execute_bash_online.php

Both these work for me with sh and bash:

arr1=("192.168.3.4" "192.168.3.4" "192.168.3.3")

strArr=$(removeDupes arr1) # strArr is a string
for item in $strArr; do arr2+=("$item"); done # convert it to an array
len2=${#arr2[@]} # get array length
echo "${len2}" # echo length

eval arr3=("$(removeDupes arr1)") # shellcheck does not like this line and won't suppress it but it works
len3=${#arr3[@]} # get array length
echo "${len3}" # echo length

As an aside, the removeDupes function looks like this:

removeDupes() {
  arg="$1[@]"
  arr=("${!arg}")
  len=${#arr[@]}
  resultArr=()
  # some array manipulation here
  echo "${resultArr[@]}"
}

This answer is based on but better explains and simplifies the answers from @Hans and @didierc

Your function doesn't return an array. It creates a global array variable called $a.

function create_some_array(){
  for i in 0 1 2 3 .. 10
  do
    a[i]=$i
  done
}

To return an array:

function create_some_array(){
  local a
  for i in 0 1 2 3 .. 10; do
    a[i]=$i
  done
  printf "%s\n" "${a[@]}"
}

# capture the created array:
readarray -t my_array < <(create_some_array)

# print its elements
for element in "${my_array[@]}"; do
  echo "  - ${element}"
done

This method, doesn't just simulate a returned array, but it also respects the spaces in a particular element.

Hope it helps

Edit: To handle returning an empty array

If you have some logic in your function that requires you to return an empty array, the above will cause an array with a single element \n to be returned. You may not want this.

To get around this, you can simply wrap the printf around a check for if the array is empty like so:

function create_some_array(){
  local a
  for i in 0 1 2 3 .. 10; do
    a[i]=$i
  done
  if [[ ${#a[@]} != 0 ]]; then
    printf "%s\n" "${a[@]}"
  fi
}