How to write to array in Bash [duplicate]

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
– glenn jackman, Commented May 5, 2019 at 13:24
@cyrus, in the recursive permutation call within the function. — glenn jackman
– glenn jackman, Commented May 5, 2019 at 13:25

Charles Duffy · Accepted Answer · 2019-05-05 16:12:55Z

1

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

answered May 5, 2019 at 16:12

community wiki

Charles Duffy

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

How to write to array in Bash [duplicate]

1 Answer 1

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Linked

Related