Create variables and assign values via loop (bash)

Question

Is there a way to create bash variables and assign them values via a loop?

Something along the lines of:

#!/bin/bash

c=0
for file in $( ls ); do
    var"$c"="$file";
    let c=$c+1;
done

EDIT: Thank you to @Costas and @mdpc for pointing out this would be a poor alternative to a list; the question is theoretical only.

Answer 1

Well, you absolutely can using eval as follows:

c=0
for file in $( ls ); do
    eval "var$c=$file";
    c=$((c+1));
done

This code will create variables named var0, var1, var2, ... with each one holding the file name. I assume you will have a good reason you want to do that over using an array ...

Answer 2

Possible you mean array. There are some ways to assign values

First:

c=0
for file in $( ls ); do
    var[$c]="$file";
    c=$(($c+1));
done

Second:

c=0
for file in $( ls ); do
    var[c++]="$file";
done

Third:

for file in $( ls ); do
    var[${#var}]="$file";
done

Fourth:

var=( $(ls) )

Fifth

...

1. there is no need to use ls , just put * for all files
2. inside [] the $ sign can be ommited