how to pass array into function and updates to array are reflected outsite function

Question

I am trying to pass a array into a function and whatever changes made to the array is reflected outside the function

function update_array()
{   
    ${1[0]}="abc" # trying to change zero-index array to "abc" , 
                  #  bad substitution error


}

foo=(foo bar)

update_array foo[@]

for i in ${foo[@]}
    do
       echo "$i" # currently changes are not reflected outside the function

    done

My questions are

1) How do i access the index array eg: zero index array , in the function , what is the syntax for it

2) How do i make changes to this index array so that the changes are reflected outsite the function also

Answer 1

Several problems, in logical order of fixing:

• Style (pet peeve)

• With your ${...} statement in update_array(), the ${..} syntax to use a variable, not define it.

  Example:

  foo[0]=abc  # assigns 'abc' to foo[0]
  

• Working around that the array name is stored in a variable.

  Not working:

  $1[0]=abc
  

  Working:

  declare -g "$1[0]=abc"  # -g arg is for a global variable
  

• Passing an argument to update_array() should pass the variable name (foo in this case), not the contents of the array. foo[@] is nothing special, it is a completely normal string (in Bash).

• Variable expansion with ${foo[@]} should be double-quoted.

Working version of the code is below:

update_array() {   
    declare -g "$1[0]=abc"
}

foo=(foo bar)

update_array foo

for i in "${foo[@]}"; do
    echo "$i"
done

## Following line added by me for further clarification
declare -p foo

which prints, correctly:

abc
bar
declare -a foo='([0]="abc" [1]="bar")'

Answer 2

Declaration of variables might not be necessary in Bash.¹ Yes you can use declare/typeset for more control over your bash variables. So I reckon you don't have to create a function just for the purpose of declaring a new array.

This script below demonstrates a direct defining of array:

#!/bin/bash
function define_array_elements() {
# You can note the array elements being defined directly, without a prior 
# definition of the variable.
for i in {1..10}; do
    var_name[$i]="Field $i of the list"
done
}

define_array_elements > /dev/null

for i in {1..10}; do
    echo "Field $i is: ${var_name[$i]}"
done

(An example borrowed from How to declare an array but not define it? with just a little modification.)

Answer 3

Short answer is: you cannot. Bash does not have a way to pass variables by reference, so there's no generic way of doing it; you're left with (ugly) hacks including indirection and/or eval.

The upcoming bash 4.3 will introduce nameref variables, which allow you to pass variables by reference, but even this feature falls short since you still risk name collition.

# example of passing variables by reference in bash 4.3
update_array() {
    declare -n array=$1
    array[0]=abc
}

foo=( foo bar )

update_array foo

printf '<%s>\n' "${foo[@]}" # outputs <abc> and <bar>

In that example, if the array was named array instead of foo, it would fail since declare -n array=array is an error (declare: array: nameref variable self references not allowed).

See http://mywiki.wooledge.org/BashFAQ/006 for other hacks.