Preserve Quotes in bash arguments

The quotes are removed before the arguments are passed to your script, so it's too late to preserve them. What you can do is preserve their effect when passing the arguments to the inner command, and reconstruct an equivalent quoted/escaped version of the arguments for printing.

For passing the arguments to the inner command "$@" -- with the double-quotes, $@ preserves the original word breaks, meaning that the inner command receives exactly the same argument list that your script did.

For printing, you can use the %q format in bash's printf command to reconstruct the quoting. Note that this won't always reconstruct the original quoting, but will construct an equivalent quoted/escaped string. For example, if you passed the argument 'uptime ; uname -a' it might print uptime\ \;\ uname\ -a or "uptime ; uname -a" or any other equivalent (see @William Pursell's answer for similar examples).

Here's an example of using these:

printf "Running command:"
printf " %q" innercmd "$@" # note the space before %q -- this inserts spaces between arguments
printf "\n"
innercmd "$@"

If the user invokes your command as:

./script 'foo'

the first argument given to the script is the string foo without the quotes. There is no way for your script to differentiate between that and any of the other methods by which it could get foo as an argument (eg ./script $(echo foo) or ./script foo or ./script "foo" or ./script \f\o""''""o).

If you want to print the argument list as close as possible to what the user probably entered:

#!/bin/bash
chars='[ !"#$&()*,;<>?\^`{|}]'
for arg
do
    if [[ $arg == *\'* ]]
    then
        arg=\""$arg"\"
    elif [[ $arg == *$chars* ]]
    then
        arg="'$arg'"
    fi
    allargs+=("$arg")    # ${allargs[@]} is to be used only for printing
done
printf '%s\n' "${allargs[*]}"

It's not perfect. An argument like ''\''"' is more difficult to accommodate than is justified.