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

Solution 1:

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

#!/bin/bash
function 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/bash
function 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"

Solution 2:

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 2
zo

It also works for associative arrays:

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

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

Solution 3:

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=NOEFLDJF

f "$b" "${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[@]}"