Variable as bash array index?

Question

#!/bin/bash

set -x

array_counter=0
array_value=1

array=(0 0 0)

for number in ${array[@]}
do
    array[$array_counter]="$array_value"
    array_counter=$(($array_counter + 1))
done

When running above script I get the following debug output:

+ array_counter=0
+ array_value=1
+ array=(0 0 0)
+ for number in '${array[@]}'
+ array[$array_counter]=1
+ array_counter=1
+ for number in '${array[@]}'
+ array[$array_counter]=1
+ array_counter=2
+ for number in '${array[@]}'
+ array[$array_counter]=1
+ array_counter=3

Why does the variable $array_counter not expand when used as index in array[]?

What's it not doing that you expect? After your loop, array is "1 1 1" — evil otto
– evil otto, Commented Aug 8, 2012 at 0:51
I would expect the debug output to show array[0]=1, array[1]=1 and so on. — John Hansen
– John Hansen, Commented Aug 8, 2012 at 0:57
ah, you're just worried about the debug output. I'm actually surprised it works at all without an eval, but it could be that lhs arrays are expanded differently. — evil otto
– evil otto, Commented Aug 8, 2012 at 1:12

Bob the Angry Coder · Accepted Answer · 2012-08-08 00:59:56Z

35

Bash seems perfectly happy with variables as array indexes:

$ array=(a b c)
$ arrayindex=2
$ echo ${array[$arrayindex]}
c
$ array[$arrayindex]=MONKEY
$ echo ${array[$arrayindex]}
MONKEY

answered Aug 8, 2012 at 0:59

Bob the Angry Coder

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1 Comment

Adrian Frühwirth Over a year ago

...and since array indices get evaluated in an arithmetic context you don't even need $, you can even do things like ${array[arrayindex-1]} which validly refers to b.

user2038893 · Accepted Answer · 2014-04-01 11:20:41Z

-2

Your example actually works.

echo ${array[@]}

confirms this.

You might try more efficient way of incrementing your index:

((array_counter++))

edited Apr 1, 2014 at 11:20

answered Apr 1, 2014 at 11:11

user2038893

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