Bash, create multiple arrays with different names using loops

Question

I'm currently learning script shell programming(bash), and trying to make multiple arrays with different names by looping so that I could put the data according to the length of my data files. I've done getting the length of the data file(marked as 'num'), and here's what I could think in my brain.

for (( i=0; i<num; i++ ))
do  
    let array$i=()
done

I've looked up many method but it didn't work. I'm a newbie to bash so any comments / recommendations / teachings would be so appreciated. Thank you.

What you are looking for is probably namerefs (with a recent enough version of bash), or indirection. Just type man bash and then search for one or the other. Note: let is not what you apparently think. In bash it introduces arithmetic expressions. — Renaud Pacalet
– Renaud Pacalet, Commented Feb 15, 2022 at 12:32
@RenaudPacalet : How can a nameref be used to dynamically create a set of arrays? I only know how we can use it to refer to an existing variable, i.e. we have manually created a number of arrays and use a nameref to dynamically refer to one of them. The OPs problem seems to be to actually create the arrays, without explicitly naming them. Of course, an eval arr$i=(....) would work, but is ugly and error-prone. — user1934428
– user1934428, Commented Feb 15, 2022 at 13:43
See Creating a string variable name from the value of another string, How can I generate new variable names on the fly in a shell script?, and Dynamic variable names in Bash. — pjh
– pjh, Commented Feb 15, 2022 at 13:46
Does this answer your question? Dynamic variable names in Bash — pjh
– pjh, Commented Feb 15, 2022 at 13:47
@user1934428 Easy. Always use the same nameref but let it point to different variable names: declare -n nr=array1, then do whatever you want with variable nr, it will be the same as if you do it with variable array1. Example: nr=([foo]=1 [bar]=2); printf '%s\t%s\n' "${!array1[@]}" "${array1[@]}" will print foo\tbar\n1\t2\n. Give it a try. Next you can declare -n nr=array2 and start working with your second array. And of course you can declare -n nr="array$i" if you wish. — Renaud Pacalet
– Renaud Pacalet, Commented Feb 15, 2022 at 14:25

Renaud Pacalet · Accepted Answer · 2022-02-16 07:26:44Z

2

With a recent enough bash, using namerefs:

for (( i=0; i<num; i++ )); do  
  declare -n nr="array$i"
  declare -a nr=()
done

And then, each time you want to add elements to your array10:

i=10
declare -n nr="array$i"
nr+=( foo bar )

To access the elements you can use the nameref (nr) or the array name, it works the same (as long as you do not modify the nameref attribute):

$ i=10
$ declare -n nr="array$i"
$ printf '%s\n' "${nr[@]}"
foo
bar
$ printf '%s\n' "${array10[@]}"
foo
bar

Think of it as a kind of reference.

edited Feb 16, 2022 at 7:26

answered Feb 15, 2022 at 14:30

Renaud Pacalet

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