Fill array passed as argument in bash function

Note that many distributions are just starting to ship with bash 4.2, so 4.3 may be a long wait if you don't want to upgrade yourself.

ksh has had nameref variables for some time. Nice to see bash catching up for this very useful feature.

It will be a great answer, once you can reasonably expect someone to be using bash 4.3 :)

@glennjackman Indeed. I need to spend some time learning about ksh, if only to know what all bash has borrowed :)

Community · Accepted Answer · 2017-05-23 11:59:43Z

0

In this answer using such technique. Modified, you can

addtoarray () { var="$1"; shift 1; eval "$var+=($(printf "'%s' " "$@"))"; }

arr=('my item 0' 'my item 1')
printf "%s\n" "${arr[@]}"
echo ====
addtoarray arr 'my new item 2' 'my new item 3'
printf "%s\n" "${arr[@]}"

prints

my item 0
my item 1
====
my item 0
my item 1
my new item 2
my new item 3

works also for initializing arrays

addtoarray secarr $(seq 5)
printf "%s\n" "${secarr[@]}"

prints:

EDIT

Decomposition of the function - works as code generator

addtoarray () { var="$1"; shift 1; eval "$var+=($(printf "'%s' " "$@"))"; }

the function's 1st argument is the variable name(!!!) (not value)
storing the name (in $1) it into $var and shifting it out from the arguments
now the arguments contains only new elements for the array like val1 val2
now constructing an bash command:

somearrayname+=('elemnts1' 'element2' .... )

where the somearrayname is the variable name what we passed as the 1st arg
the printf "'%s '" "$@" creates the single-quoted array members from the arg-list
so in case of calling the function as addtoarray arr val1 "spaced val2" we generating the next string

arr+=('val1' 'spaced val2' )

this is an correct construction to adding members to the array named arr - regardless if before contets, e.g. adds the new elements to its end. (if it was empty, the end is its start)

finally the eval executes the above generated string
in the result you got an initialized/modified array $arr (the $arr is global variable, therefore is visible in the function too)

And finally - brutally stole @chepner's answer ;) using his technique the addtoarray si simple as:

addtoarray () {
    declare -n arrname=$1
    shift 1;
    arrname+=("$@")
}

If you have bash 4.3 you should accepts @chepner's answer - it is the best.

edited May 23, 2017 at 11:59

CommunityBot

11 silver badge

answered Sep 5, 2014 at 10:29

clt60

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5 Comments

Your code does the trick. But what's happening? In main the array is filled with the new value. In the function, var seems to only carry the main variable name (it's not an array, like I thought).

clt60 Over a year ago

@il_mix added the explanation

Obligatory warning: eval will execute anything you pass as a parameter; it doesn't care if the generated string winds up being more than a simple array assignment

clt60 Over a year ago

@chepner - yes, of course - in this case before the eval all passed arguments are stored into single quoted strings what helps a bit... ;) (should sanitize the array name tough)...

Fravadona Over a year ago

Please replace eval "$var+=($(printf "'%s' " "$@"))"; with eval "$var+=($(printf '%q ' "$@"))";

anubhava · Accepted Answer · 2014-09-05 10:16:15Z

0

You are adding elements into an array variable inside the function but arrays or any other variables are not passed as reference in BASH. So any changes made in those variables inside the function won't be visible outside the function.

As a workaround you can use a global variable like this example:

# initialize an array
myArray=("my item 0" "my item 1")

useArrayInFunc { myArray+=("my item 2")); }

useArrayInFunc

printf "%s\n" "${myArray[@]}"
my item 0
my item 1
my item 2

answered Sep 5, 2014 at 10:16

anubhava

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2 Comments

I prefer not to use a global variable

Since I need to use the function on different arrays, and then do some operations with these arrays, I need to call it with different parameters. Otherwise, I can let the function operate on a global variable, and after the function call copy the global array values to the actual array; this is kind of tedious...

glenn jackman · Accepted Answer · 2014-09-05 10:26:57Z

0

You need to assign the local var to the global one:

function fillArray {
  local arrayName=$1
  local ref=$arrayName[@]
  local array=("${!ref}")
  local i item key

  for i in 0 1 2 3
  do
    array+=("new item $i")
  done

  echo "Tot items in function: ${#array[@]}"
  for item in "${array[@]}"
  do
    echo $item
  done
  echo

  for key in "${!array[@]}"; do
    declare -g ${arrayName}["$key"]="${array[$key]}"
  done
}

answered Sep 5, 2014 at 10:26

glenn jackman

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5 Comments

I tried your example, but I get this error: "declare: -g: invalid option". I'm using bash 4.1.2

Bummer. Working with arrays in bash is such a PITA. You might want to change your approach: call the function to generate the new items to add to the array, but return (via 'echo') a newline/null-byte delimited string. Add the elements to the array in the current scope, not in the function.

confirmed that -g was added to declare in version 4.2

bash 4.2 is 3 years old now. Any reason you can't upgrade?