Adding value to array passed as an argument doesn't add value to it

Question

When I try to pass an array as parameter and add a value to it, it end's up as being empty at the end.

How would I add value to passed array? I would make the array global, but I will be passing different arrays, so it must be passed to this function as parameter.

#!/bin/bash

base='/home'

declare -a files_new
declare index

arrayit() {
    #$1 = path
    #$2 = array

    dir=($1/*)
    arr=$2

    for directory in "${dir[@]}"
    do
        if [[ -d $directory ]]
        then
            arrayit $directory $arr
        else
            arr[$index++]=$directory
        fi
    done
}

index=0
arrayit $base $files_new

for file in "${files_new[@]}"
do
    echo "File: $file"
done

I assume. But I'm not sure if I'm declaring it correctly inside function with $2. — HelpNeeder
– HelpNeeder, Commented Feb 6, 2016 at 19:18
arr is not an array; you are passing the first element ($files_new is equivalent to ${files_new[0]}) as a regular string, and assigning that to arr. — chepner
– chepner, Commented Feb 6, 2016 at 19:38
Thanks chepner; Would passing ${files_new[@]} be considered as passing entire array? — HelpNeeder
– HelpNeeder, Commented Feb 6, 2016 at 19:42
It would pass the contents of the array, not the array itself (meaning you wouldn't be able to modify the array from within the function). — chepner
– chepner, Commented Feb 6, 2016 at 19:48
If you are using bash 4, you can throw out arrayit altogether and just set files_new=( "$base"/**/*/ ) to get a recursive list of directory names. (With shopt -s globstar to enable use of **.) — chepner
– chepner, Commented Feb 6, 2016 at 19:58

glenn jackman · Accepted Answer · 2016-02-06 21:52:57Z

1

Passing around arrays in bash is a real PITA. I've found this is the most reliable method

$ ary=(foo bar baz)
$ fn() { local tmp="${1}[@]"; local copy=("${!tmp}"); declare -p copy; }
$ fn ary
declare -a copy='([0]="foo" [1]="bar" [2]="baz")'

That gives you a copy in the function. Changes made in the function will not appear in the parent scope.

However, bash version 4.3 has "name references"

$ fn() { 
    local -n myary=$1       # note: -n -- see "help declare"
    declare -p myary
    for key in "${!myary[@]}"; do printf "%s\t%s\n" "$key" "${myary[$key]}"; done
    myary[42]="new value"
}
$ fn ary
declare -n myary="ary"
0   foo
1   bar
2   baz
$ declare -p ary
declare -a ary='([0]="foo" [1]="bar" [2]="baz" [42]="new value")'

Note the new value in the parent scope! Hooray!

answered Feb 6, 2016 at 21:52

glenn jackman

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1 Comment

glenn jackman Over a year ago

Note that in both cases, I'm passing the name of the array to the function.

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