Named bash function parameters

Question

Always wondered this, but never fully investigated - is there any way to get named parameters in bash?

For example, I have this:

function ql_maybe_fail {
  if [[ "$1" == "true" ]]; then
      echo "quicklock: exiting with 1 since fail flag was set for your 'ql_release_lock' command. "
      exit 1;
  fi
}

is it somehow possible to convert it to something like this:

function ql_maybe_fail (isFail) {
  if [[ "$isFail" == "true" ]]; then
      echo "quicklock: exiting with 1 since fail flag was set for your 'ql_release_lock' command. "
      exit 1;
  fi
}

Or https://unix.stackexchange.com/questions/122632/a-bash-function-that-takes-argument-like-other-languages — muru, Mar 02 '18 at 03:23
damn those answers are all so unclear...after reading, I guess the answer is a big fat no, lol — Alexander Mills, Mar 02 '18 at 03:30
The best I've found is this answer over on StackOverflow - https://stackoverflow.com/a/7948533 — Chris Davies, Mar 02 '18 at 08:39
What's the problem that you are trying to solve that requires this to work? — Kusalananda, Mar 02 '18 at 10:29
it's just for convenience - most programming languages have named arguments, etc, for a reason — Alexander Mills, Mar 03 '18 at 20:57

score 6 · Accepted Answer · answered Mar 02 '18 at 04:13

6

Functions in Bash currently do not support user-named arguments.

answered Mar 02 '18 at 04:13

fpmurphy

4,636

score 4 · Answer 2 · answered Mar 02 '18 at 09:57

4

This workaround might help, but it is not well testet:

fun () {
    v1=$1
    v2=$2
    for v in "$v1" "$v2"
    do
       case "$v" in
           name=*) name=${v/*=/};;
           age=*)  age=${v/*=/};;
           *)    echo "unexpected $v, please use name and age" ;;
       esac
    done

    echo "name=$name age=$age"
}

output:

fun "name=John" "age=22"
name=John age=22
fun "age=22" "name=John"
name=John age=22

answered Mar 02 '18 at 09:57

user unknown

10,482

yeah, it's just not the same lol – Alexander Mills Mar 02 '18 at 10:03
3

Well, for your sample (function ql_maybe_fail (isFail) {) just write function ql_maybe_fail () { isFail=$1 - where is the problem? – user unknown Mar 02 '18 at 10:06
no problem just an extra line of code I guess – Alexander Mills Mar 03 '18 at 20:57

Named bash function parameters

2 Answers2