Bash: How to declare empty array, and then add variables to it

Question

I was hoping a kind person more intelligent than me could help me out here.

I am working on a Bash script, and in it there is a for loop that will go around an unknown/undefined number of times.

Now, in this for loop, there will be a value assigned to a variable. Let's call this variabe: $var1

Each time the loop goes (and I will never know how many times it goes), I would like to assign the value inside $var1 to an array, slowly building up the array as it goes. Let's call the array $arr

This is what I have so far:

for i in $( seq 0 $unknown ); do
    ...
    some commands that will make $var1 change...
    ...

    arr=("${arr[@]}" "$var1")
done

However, when I want to echo or use the values in the array $arr, I get no results

Perhaps someone will kindly help me in the right direction?

I would really appreciate it so much.

arr+=("$var1") will populate the array inside the loop, and if you really want to define an empty array before/outside the loop, arr=() Or declare -a arr — Jetchisel
– Jetchisel, Commented Mar 30, 2020 at 23:07
Show us how you're trying to use the values in the array, and ensure that you're providing not pseudocode but a working example we can run ourselves without needing to make any changes to see your problem (as described in the minimal reproducible example definition). — Charles Duffy
– Charles Duffy, Commented Mar 30, 2020 at 23:22
BTW, in general, for i in $(seq 0 $unknown) would be better written as for ((i=0; i<unknown; i++)). seq is not a standardized command or part of the shell itself, and running an external command when not needed is generally inefficient. (The POSIX-compliant alternative is wordier: i=0; while [ "$i" -lt 10 ]; do ...; i=$((i + 1)); done -- there's a reason ksh and bash provide the shorter form). — Charles Duffy
– Charles Duffy, Commented Mar 30, 2020 at 23:24
I've voted to re-open this - it appears to me that, while the proposed dupe was related to arrays, it was a totally different question regarding why echo ${arr} only ever gave the first element. — paxdiablo
– paxdiablo, Commented Mar 30, 2020 at 23:29

paxdiablo · Accepted Answer · 2025-06-12 13:51:57Z

8

You declare and add to a bash array as follows:

declare -a arr       # or arr=()
arr+=("item1")
arr+=("item2")

After executing that code, the following assertions (among others) are true:

${arr[@]}  = item1 item2
${#arr[@]} = 2
${arr[1]}  = item2

In terms of the code you provided, you would use something like:

declare -a arr
for i in $( seq 0 $unknown ); do
    ...
    some commands that will make $var1 change...
    ...
    arr+=("$var1")
done

edited Jun 12 at 13:51

answered Mar 30, 2020 at 23:09

paxdiablo

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