How to write to array in Bash

Question

Does somebody know what is wrong and why it doesn't want to add all permutations in function into an array "array" and when I run for loop only gives me last value

#!/bin/bash
array=()
permutation() {

  local items="$1"
  local out="$2"
  local i
  [[ "$items" == "" ]] && array[$i]="$out" && return
  for (( i=0; i<${#items}; i++ )) ; do
    permutation "${items:0:i}${items:i+1}" "$out${items:i:1}"
  done
  }

permutation $1

for i in "${array[$i]}"
do 
  echo "$i"
done

You re-initialize the global `array` variable every time you enter the function. Move `array=()` outside of the function. — glenn jackman, May 05 '19 at 13:24
@cyrus, in the recursive `permutation` call within the function. — glenn jackman, May 05 '19 at 13:25

score 0 · Answer 1 · answered May 05 '19 at 16:12

Use array+=( "$out" ) to append to the end of your array without needing to know the index of that value. Because you're declaring i local, it isn't shared across the invocations of your functions, so they're all overwriting ${array[0]}.
for i in "${array[$i]}"; do ignores all the values of your array except for whichever one is in position $i, for the single value of i that exists before your for loop is entered. Use for i in "${array[@]}" to iterate over all values in the array.

See a corrected version of your code running at https://ideone.com/zIYigA

How to write to array in Bash

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