Bash - Passing an Array to a Function

#!/bin/bash : ' VERSION 1 Loop through array using variable name ' loopArray () { echo "First argument passed: "$1; local arrayname=$1 # check for named arguments [-a || --array] for i in "$@" ; do case $i in -a=*|--array=*) a="${i#*=}" local arrayname=$a ;; esac done # turn arrayname into local array local tmp=$arrayname[@] local array=( "${!tmp}" ) # loop through values for value in "${array[@]}" ; do echo $value done } : ' define external list ' LIST=( 'abc' '123' 'The quick brown fox jumps over the lazy dog' ) : ' Test ' result=$(loopArray LIST); echo $result loopArray LIST loopArray -a=LIST loopArray --array=LIST : ' First argument passed: LIST abc 123 The quick brown fox jumps over the lazy dog First argument passed: LIST abc 123 The quick brown fox jumps over the lazy dog First argument passed: -a=LIST abc 123 The quick brown fox jumps over the lazy dog First argument passed: --array=LIST abc 123 The quick brown fox jumps over the lazy dog ' : ' VERSION 2 Function accepts array values as first param ' loopArray () { echo "First argument passed: "$1; # convert the first argument # from scalar to new array # NOTE: whitespace space acts as the delimiter _array=($1) # check for named arguments [-a || --array] for i in "$@" ; do case $i in -a=*|--array=*) a="${i#*=}" _array=($a) ;; esac done # loop through values for value in "${_array[@]}" ; do echo $value done } : ' define external list ' list=( 'a' 'b' 'c' 'two words' # this method won't work with spaces ) : ' Test Must pass the array as scalar values -- no spaces ' result=$(loopArray "${list[*]}"); echo $result loopArray "${list[*]}" loopArray -a="${list[*]}" loopArray --array="${list[*]}" : ' First argument passed: a b c two words a b c two words First argument passed: a b c two words a b c two words First argument passed: -a=a b c two words a b c two words First argument passed: --array=a b c two words a b c two words '