bash/shell pass array to function then loop?

Question

When defining an Array directly in the function the below works as expected;

function each {
  local array=(1 2 3)
  len=${#array[*]}
  for (( i=0; i<len; i++ )); do
    echo ${array[$i]}
  done
}

each

# outputs;

1
2
3

But I can't figure out how to pass the array in as an argument and achieve the same output;

function each {
  local array=$1
  len=${#array[*]}
  for (( i=0; i<len; i++ )); do
    echo ${array[$i]}
  done
}

array=(1 2 3)
each array

# outputs;

array

Grateful for any help, thanks.

score 1 · Answer 1 · answered Feb 24 '13 at 14:07

1

I would pass the array elements as distinct positional parameters to the function:

function f() {
    local array = "$@";
    # ...
}

## calling your function
f ("${array[@]}")

answered Feb 24 '13 at 14:07

Costi Ciudatu

37,042
7
56
92

thanks Costi, will close this as dupe though – Jamie Mason Feb 24 '13 at 14:39
1

I voted for closing it myself after realizing there's a complete discussion already clarifying this. :) – Costi Ciudatu Feb 24 '13 at 15:22

bash/shell pass array to function then loop?

1 Answers1