Function to iterate over array

Question

I am using the following script, so as to call a function that is supposed to iterate over an array.

#!/bin/bash function iterarr { for item in "$1" do echo "$item" done } myarr=(/dir1/file1.md /dir1/file2.md README.md) iterarr "${myarr[@]}" 

However, when I execute it, it gives the following output.

/dir1/file1.md

Why does it print only the first array entry?

edit: What is more, I would like to be able to use an additional argument (besides the array, so if I use '$@', how to I access the second argument?)

Working on Ubuntu 16.04.03 with ...

*$ $(which bash) --version GNU bash, version 4.3.48(1)-release (x86_64-pc-linux-gnu) Copyright (C) 2013 Free Software Foundation, Inc. License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html> 

Answer 1

iterarr "${myarr[@]}" will expand to iterarr '/dir1/file1.md' '/dir1/file2.md' 'README.md' and in your loop you only reference the first argument with "$1". Instead use "$@" to loop over all of the arguments.

#!/bin/bash function iterarr { for item in "$@" do echo "$item" done } myarr=(/dir1/file1.md /dir1/file2.md README.md) iterarr "${myarr[@]}" 

If you want to process flags, or positional arguments then place them before the array and handle them first, shifting them when done will remove them from "$@",

#!/bin/bash function iterarr { echo "first argument is : '$1'" shift for item in "$@" do echo "$item" done } myarr=(/dir1/file1.md /dir1/file2.md README.md) iterarr firstarg "${myarr[@]}" 

Answer 2

With this current version of bash, use a "nameref" to pass the array by name:

iterarr() { local -n local_arr=$1 for item in "${local_arr[@]}"; do echo "$item"; done } myarr=(/dir1/file1.md /dir1/file2.md README.md) iterarr myarr 

/dir1/file1.md /dir1/file2.md README.md