bash/shell pass array to function then loop? [duplicate]

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.

Costi Ciudatu · Accepted Answer · 2013-02-24 14:07:13Z

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, 2013 at 14:07

Costi Ciudatu

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2 Comments

Jamie Mason Over a year ago

thanks Costi, will close this as dupe though

Costi Ciudatu Over a year ago

I voted for closing it myself after realizing there's a complete discussion already clarifying this. :)

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