common bash function to check arguments and if "--help" was passed, then call a help function

Question

I have the following condition in a few bash functions to print their help message, but I realized that this is such an annoying repetition and wonder if there's a way around it so that the condition and call happen in a single wrapper function. So instead of this:

fn1() {
  arg=$@
  [ ! -z $arg ] || [ $arg = "--help" ] && helpit "fn1 help msg." && return

  # fn body here
}
fn2() {
  arg=$@
  [ ! -z $arg ] || [ $arg = "--help" ] && helpit "fn2 help msg." && return

  # fn body here
}

We have this:

try_help() {
  # FYI $@ is the help msg and not the params of its caller
  arg=$@
  [ ! -z $arg ] || [ $arg = "--help" ] && helpit "Delete a remote branch by name." && return 1
  return 0
}
fn1() {
  try_help "fn1 help msg." || return

  # fn body here
}
fn2() {
  try_help "fn2 help msg." || return

  # fn body here
}

Is this possible to do?

arg=$@ doesn't work the way you want, btw. It's exactly identical to arg=$*, because $@ can't be stored in a string variable at all (it can only be stored in arrays). — Charles Duffy
– Charles Duffy, Commented Apr 3, 2022 at 12:44
@CharlesDuffy so we can't forward them to the try_help for checking? — Jackson
– Jackson, Commented Apr 3, 2022 at 12:45
You can absolutely forward them, you just can't do it with a string variable. You could use a global/shared array, or you can just pass them on try_help's argument list. — Charles Duffy
– Charles Duffy, Commented Apr 3, 2022 at 12:47
BTW, ... && helpit "something" && return 1 is a bad idea, because it executes return 1 only if helpit has a successful exit status. Better to run ... && { helpit "something"; return 1; } so the return 1 runs even if helpit reports a failure. — Charles Duffy
– Charles Duffy, Commented Apr 3, 2022 at 12:51
...to go more into why [ -z $arg ] is buggy: When you have an unquoted empty argument list, it turns into no arguments at all, so you get [ -z ], which is treated like [ -n "-z" ], which is why it appears to work. But it only "works" when $arg is a simple enough value; consider arg='foo bar'; you get [ -z "foo" "bar" ], which is an invalid test command so the code throws an error instead of just silently returning false as it should. — Charles Duffy
– Charles Duffy, Commented Apr 3, 2022 at 12:55

Charles Duffy · Accepted Answer · 2022-04-03 12:46:53Z

3

Trivially done.

helpit() { echo "$*" >&2; }
try_help() {
  local help_msg arg
  help_msg=$1; shift
  if (( $# == 0 )); then
    echo "No arguments found. Printing help:" >&2
    helpit "$help_msg"
    return 1
  fi
  for arg in "$@"; do
    case $arg in
      --)     return 0;;
      --help) helpit "$help_msg"; return 1;;
    esac
  done
  return 0
}

fn1() {
  try_help "fn1 help msg" "$@" || return
  # fn body here
}
fn2() {
  try_help "fn2 help msg" "$@" || return
  # fn body here
}

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