[SOLVED] Detecting invalid options

Faki · 04-27-2023, 02:34 PM

I process positional parameters in the following way. I use the pattern `(-*)` for detecting invalid options - is this good or can one be more precise?

Code:

  local parg=""
  while (( $# > 0 )); do
    parg="$1"
    case $parg in
      ##------------------------------------------------
      (-p+([[:digit:]]))
          val="${parg:2}"    
          shift 1
          ;;
      ##------------------------------------------------
      ("--")
          break ;;
      (-*)
          printf '%s\n' "$parg Unrecognised Predicate"
          break ;;
      (*)
          break ;;
    esac
  done

MadeInGermany · 04-28-2023, 02:42 AM

It's okay.
I would place the shift before the done so it's immediately visible that the loop won't run forever.
And shift out all -* args.

Code:

  while (( $# > 0 )); do
    case $1 in
    ##------------------------------------------------
    (-p+([[:digit:]]))
      val="${1:2}" ;;
    (-p*)
      printf '%s\n' "${1:2} malformed number" ;;
    (--)
      shift; break ;;
    (-*)
      printf '%s\n' "$1 Unrecognised Predicate"
      shift; break ;;
    (*)
      break ;;
    esac
    shift
  done

Faki · 04-28-2023, 09:00 AM

Interesting. You do this on a regular basis? In many instances others shift the required amount within the case. I wonder if - on its own should behave the same as --. Is it correct to say that (-*) should be described by "Unrecognised Predicate" ?

MadeInGermany · 04-28-2023, 10:29 AM

Some old Unix commands have a single - as a "last option" indicator.
But Posix says it must be --

Faki · 04-29-2023, 04:35 PM

Quote:

Originally Posted by MadeInGermany

It's okay.
I would place the shift before the done so it's immediately visible that the loop won't run forever.
And shift out all -* args.

Code:

  while (( $# > 0 )); do
    case $1 in
    ##------------------------------------------------
    (-p+([[:digit:]]))
      val="${1:2}" ;;
    (-p*)
      printf '%s\n' "${1:2} malformed number" ;;
    (--)
      shift; break ;;
    (-*)
      printf '%s\n' "$1 Unrecognised Predicate"
      shift; break ;;
    (*)
      break ;;
    esac
    shift
  done

I do not see that it shifts out all -* args. Because on the first one, you exit the loop.