Convert command line arguments into an array in Bash

Question

How do I convert command-line arguments into a bash script array?

I want to take this:

./something.sh arg1 arg2 arg3

and convert it

Answer 1

The importance of the double quotes is worth emphasizing. Suppose an argument contains whitespace.

Code:

#!/bin/bash
printf 'arguments:%s\n' "$@"
declare -a arrayGOOD=( "$@" )
declare -a arrayBAAD=(  $@  )

printf '\n%s:\n' arrayGOOD
declare -p arrayGOOD
arrayGOODlength=${#arrayGOOD[@]}
for (( i=1; i<${arrayGOODlength}+1; i++ ));
do
   echo "${arrayGOOD[$i-1]}"
done

printf '\n%s:\n' arrayBAAD
declare -p arrayBAAD
arrayBAADlength=${#arrayBAAD[@]}
for (( i=1; i<${arrayBAADlength}+1; i++ ));
do
   echo "${arrayBAAD[$i-1]}"
done

Output:

> ./bash-array-practice.sh 'The dog ate the "flea" -- and ' the mouse.
arguments:The dog ate the "flea" -- and 
arguments:the
arguments:mouse.

arrayGOOD:
declare -a arrayGOOD='([0]="The dog ate the \"flea\" -- and " [1]="the" [2]="mouse.")'
The dog ate the "flea" -- and 
the
mouse.

arrayBAAD:
declare -a arrayBAAD='([0]="The" [1]="dog" [2]="ate" [3]="the" [4]="\"flea\"" [5]="--" [6]="and" [7]="the" [8]="mouse.")'
The
dog
ate
the
"flea"
--
and
the
mouse.
>