Creating n variables in Bash without assigning them one by one?

Question

How do I create n variables in shell scripts without explicitly assigning them? What I mean is something like a loop that creates var1, var2, var3,...,varx, where x is a variable I set earlier, something like:

read x

for ((a=0;a<x;++a)); do
 variable$a=${RANDOM}
done

(Let's ignore the possibility that x might be a string for now. But obviously, this doesn't work. How would one do this?

What I actually want to do, is that each argument I wrote in the command line when I executed the script to become it's own variable ARG1, ARG2... ARGn with ${1}, ${2},..., ${n} as its value, so there will only be as many of these variables, as arguments were set.

Answer 1

It looks like you want to use an array:

read x

for (( a=0; a<x; ++a)); do
   variable[a]=$RANDOM
done

printf 'First value is %s\n' "${variable[0]}"

printf 'All values (one by one): %s\n' "${variable[@]}"
printf 'All values (as one string): %s\n' "${variable[*]}"

---

For the second part of your question:

arg=( "$@" )

printf 'First command line argument: %s\n' "${arg[0]}"

Note also that you can easily loop over all command line arguments (or whatever happens to be in $@) without storing them anywhere special:

for arg do
    printf 'Got command line argument: %s\n' "$arg"
done

Answer 2

Try this,

read x

for ((a=0;a<x;++a)); do
 declare -i variable$a=${RANDOM}
done

declare command permits assigning a value to a variable in the same statement as setting its properties.

Answer 3

You need to eval the assignment (but to use an array would be better).

#!/bin/bash -vx

read x
for ((a=0;a<x;++a)); do
   eval variable$a=${RANDOM}
done

From man bash

eval [arg ...]
    The args are read and concatenated together into a single command. 
    This command is then read and executed by the shell, and its exit 
    status is returned as the value of eval.

Answer 4

How about (recent shell needed)

 for ((a=0;a<x;++a))
   do  read variable$a <<<${RANDOM}
   done