How can I escape an arbitrary string for use as a command line argument in Bash?

Question

I have a list of strings and I want to pass those strings as arguments in a single Bash command line call. For simple alphanumeric strings it suffices to just pass them verb

Answer 1

FWIW, I wrote this function that invokes a set of arguments using different credentials. The su command required serializing all the arguments, which required escaping them all, which I did with the printf idiom suggested above.

$ escape_args_then_call_as myname whoami

escape_args_then_call_as() {
    local user=$1
    shift

    local -a args
    for i in "$@"; do
        args+=( $(printf %q "${i}") )
    done

    sudo su "${user}" -c "${args[*]}"
}