How to quotes in bash function parameters?

The reason this happens is because bash interprets the arguments, as you thought. The quotes simply aren't there any more when it calls the function, so this isn't possible. It worked in DOS because programs could interpret the command line themselves, not that it helps you!

Although @Peter Westlake's answer is correct, and there are no quotes to preserve one can try to deduce if the quotes where required and thus passed in originally. Personally I used this requote function when I needed a proof in my logs that a command ran with the correct quoting:

function requote() {
    local res=""
    for x in "${@}" ; do
        # try to figure out if quoting was required for the $x:
        grep -q "[[:space:]]" <<< "$x" && res="${res} '${x}'" || res="${res} ${x}"
    done
    # remove first space and print:
    sed -e 's/^ //' <<< "${res}"
}

And here is how I use it:

CMD=$(requote "${@}")
# ...
echo "${CMD}"

I was in a similar position to you in that I needed a script to wrap around an existing command and pass arguments preserving quoting.

I came up with something that doesn't preserve the command line exactly as typed but does pass the arguments correctly and show you what they were.

Here's my script set up to shadow ls:

CMD=ls
PARAMS=""

for PARAM in "$@"
do
  PARAMS="${PARAMS} \"${PARAM}\""
done

echo Running: ${CMD} ${PARAMS}
bash -c "${CMD} ${PARAMS}"
echo Exit Code: $?

And this is some sample output:

$ ./shadow.sh missing-file "not a file"
Running: ls "missing-file" "not a file"
ls: missing-file: No such file or directory
ls: not a file: No such file or directory
Exit Code: 1

So as you can see it adds quotes which weren't originally there but it does preserve arguments with spaces in which is what I needed.