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

Question

As we know, in bash programming the way to pass arguments is $1, ..., $N. However, I found it not easy to pass an array as an argument to a function which receives more than one argument. Here is one example:

f(){
 x=($1)
 y=$2

 for i in "${x[@]}"
 do
  echo $i
 done
 ....
}

a=("jfaldsj jflajds" "LAST")
b=NOEFLDJF

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

As described, function freceives two arguments: the first is assigned to x which is an array, the second to y.

f can be called in two ways. The first way use the "${a[@]}" as the first argument, and the result is:

jfaldsj 
jflajds

The second way use the "${a[*]}" as the first argument, and the result is:

jfaldsj 
jflajds 
LAST

Neither result is as I wished. So, is there anyone having any idea about how to pass array between functions correctly?

Answer 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"