How to pass array as an argument to a function in Bash

You cannot pass an array, you can only pass its elements (i.e. the expanded array).

#!/bin/bashfunction f() {    a=("$@")    ((last_idx=${#a[@]} - 1))    b=${a[last_idx]}    unset a[last_idx]    for i in "${a[@]}" ; do        echo "$i"    done    echo "b: $b"}x=("one two" "LAST")b='even more'f "${x[@]}" "$b"echo ===============f "${x[*]}" "$b"

The other possibility would be to pass the array by name:

#!/bin/bashfunction f() {    name=$1[@]    b=$2    a=("${!name}")    for i in "${a[@]}" ; do        echo "$i"    done    echo "b: $b"}x=("one two" "LAST")b='even more'f x "$b"

You can pass an array by name reference to a function in bash (since version 4.3+), by setting the -n attribute:

show_value () # array index{    local -n myarray=$1    local idx=$2    echo "${myarray[$idx]}"}

This works for indexed arrays:

$ shadock=(ga bu zo meu)$ show_value shadock 2zo

It also works for associative arrays:

$ declare -A days=([monday]=eggs [tuesday]=bread [sunday]=jam)$ show_value days sundayjam

See also nameref or declare -n in the man page.

You could pass the "scalar" value first. That would simplify things:

f(){  b=$1  shift  a=("$@")  for i in "${a[@]}"  do    echo $i  done  ....}a=("jfaldsj jflajds" "LAST")b=NOEFLDJFf "$b" "${a[@]}"

At this point, you might as well use the array-ish positional params directly

f(){  b=$1  shift  for i in "$@"   # or simply "for i; do"  do    echo $i  done  ....}f "$b" "${a[@]}"