Bash: Save and print arguments from an array

Question

How do you save all the arguments in a bash script to an array and print them individually?

John Kugelman · Accepted Answer · 2012-07-29 17:47:35Z

6

Initialize the array:

ARGS=("$@")              # "$@" gives the arguments passed to the script
ARGS=(arg1 arg2 arg3)    # or fill the array out yourself

Display the array items:

for ARG in "${ARGS[@]}"; do
    printf '%s\n' "$ARG"
done

answered Jul 29, 2012 at 17:47

John Kugelman

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1 Comment

+1. Can also use a C-like for loop to iterate over the indices: for (( i=0; i < ${#ARGS[@]}; i++ )); do printf "%d\t%s\n" $i "${ARGS[i]}"; done