bash + for loop + output index number and element

Question

This is my array:

$ ARRAY=(one two three)

How do I print the array so I have the output like: index i, element[i] using the printf or for loop I use below

1,one
2,two
3,three

Some notes for my reference

1 way to print the array:

$ printf "%s\n" "${ARRAY[*]}"
one two three

2 way to print the array

$ printf "%s\n" "${ARRAY[@]}"
one
two
three

3 way to print the array

$ for elem in "${ARRAY[@]}"; do  echo "$elem"; done
one
two
three

4 way to print the array

$ for elem in "${ARRAY[*]}"; do  echo "$elem"; done
one two three

A nothe way to look at the array

$ declare -p ARRAY
declare -a ARRAY='([0]="one" [1]="two" [2]="three")'

Answer 1

You can iterate over the indices of the array, i.e. from 0 to ${#array[@]} - 1.

#!/usr/bin/bash

array=(one two three)

# ${#array[@]} is the number of elements in the array
for ((i = 0; i < ${#array[@]}; ++i)); do
    # bash arrays are 0-indexed
    position=$(( $i + 1 ))
    echo "$position,${array[$i]}"
done

Output

1,one
2,two
3,three

Answer 2

The simplest way to iterate seems to be:

#!/usr/bin/bash

array=(one two three)

# ${!array[@]} is the list of all the indexes set in the array
for i in ${!array[@]}; do
  echo "$i, ${array[$i]}"
done