bash - read input into variable/password in bash -

is possible store - example user input - variable in bash? have script reads password...

#!/bin/bash  while [ $# -gt 0 ];     key="$1";     case $key in         -p|--prompt)             if [ ! "$2" ];                 printf "\e31;1mmust pass in prompt\e[0m\n";                 exit 1;             fi             printf "$2";             shift 2;             ;;     esac done  psswd=; c=; while [ "$c" != "^m" ];     read -n 1 -s c;     if [ "$c" == "^?" ];         if [ ${#psswd} -gt 0 ];             psswd=${psswd:: -1};             printf "\b \b";         fi     elif [ ! "$c" ]; # pressed enter         break;     else         psswd="$psswd$c";         printf "*";     fi done;  printf "\nyour entered password is: $psswd\n" 

(sorry, ^m newline , ^? backspace. meaning can't copy , paste, how shows in vim, , don't know how else write it.)

... , password strored variable. know how read stores variable. if read can it, why can't i? have ideas?

(or perhaps there better way store password... echo "$psswd" | base64 > .pass - isn't want do.)

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update

basically, how psswd variable outside shell? or how access password once shell closed? there better way store other variable?

as rojomoke suggests, enough:

read -p "enter password: " -s pw echo "you entered: $pw" 

if this: read -n 1 -s c , hit enter, $c empty, not hold \r; , if hit backspace, in testing, $c holds character octal value 177

read -n 1 -s c printf "%s" "$c" | od -c