Passing arrays in to the function in bash script

Question

I am new to a bash script and writing the bash script. I have declared an array and I have given the values. But When I pass the array in the loop values are not getting inside the loop. Below is my code

declare -a files=( "A1S0" "D1S0" "D2S0" "D3S0" "D4S0" "D5S0" "D6S0" )
command_to_get_filesizes() {
  for i in "${!files[@]}"; do
     echo Processing file "${i[${index}]}"
  done
}
command_to_get_filesizes

Expected Output

A1S0
D1s0
D2S0
D3S0
D5S0
D680

But am getting output as

Processing file 0
Processing file 1
Processing file 2
Processing file 3
Processing file 4
Processing file 5
Processing file 6

could anyone pls help me to print values in the array

Answer 1

The variable 'index' is undefined. What you intended was

echo Processing file "${files[i]}"

What the original does is obscure. All variables are (potentially) arrays with a zeroth element. (Try echo "${PATH[$junk]}".) So that whole construct "${i[${index}]}" resolves as ${i[0]}, which just echoes $i.

Your title is slightly misleading: you do not pass the array into the function (as a parameter): it is merely a global array variable.

@Siva is correct in that you can iterate over the array values simply. However, if you need both the index and the value in the same iteration, this does it for me.

for n in "${!N[@]}"; do
    printf '%8d  %s\n' "${n}" "${N[n]}"
done

Answer 2

As per your expected output,

• we don't require "Processing file" in that echo command.
• we don't need to pass the index value of array while looping it in for loop.

Try as

declare -a files=( "A1S0" "D1S0" "D2S0" "D3S0" "D4S0" "D5S0" "D6S0" )
command_to_get_filesizes() 
{
  for i in "${files[@]}"; do
    echo "$i"
  done
}
command_to_get_filesizes