将数组直接传递给Shell中的函数

Question

is it possible to pass an array directly to a function in shell/bash, without saving it first to a variable?

currently example:

function checkParameter(){
    param1=("${!1}")
    for a in "${param1[@]}"; do
        echo "${a}"
    done

    param2=("${!2}")
    for b in "${param2[@]}"; do
        echo "${b}"
    done
}

a=( "child 1 from 1" "child 2 from 1" "child 3 from 1" )
b=( "child 1 from 2" "child 2 from 2")
checkParameter a[@] b[@]

So can I pass the ( "child 1 from 1" "child 2 from 1" "child 3 from 1" ) and the ( "child 1 from 2" "child 2 from 2") somehow directly to the function checkParameter , without saving it in a and b first?

greetings and thanks,
christopher2007

Answer 1

"$1" is immutable. That is why your example has to store the value of it in a temporary variable. This holds true for $@ regardless of whether they are passed as an array or not.

Answer 2

Are you sure that you need to use an array ?

checkparameters() {
    for param in "$@"; do
        echo "$param"
    done
}

checkparameters "a" "b" "c"

To answer the question, I don't think that is possible to pass an array as function's args without transformations.

EDIT

You could try something like :

[ ~]$ cat test.sh 
#!/bin/bash 

declare -a a1
a1=("a" "b")

declare -a a2
a2=("c" "d")

checkParameters(){
    for i in "$@"; do
        echo "Content of $i :"
        for j in $(eval "echo \${$(echo $i)[@]}"); do
            echo $j
        done
    done
}

checkParameters "a1" "a2"
[ ~]$ ./test.sh 
Content of a1 :
a
b
Content of a2 :
c
d

Answer 3

Generally, you cannot pass arrays to Bash scripts or functions as arrays :

When you pass an array on the command line, eg, "${ary[@]}" , you're not passing an array , but its elements as individual parameters ; the first array element is bound to special parameter $1 , the second to $2 , ...

Your solution attempt employs a clever workaround by passing the name of array variables to the function, where variable indirection ( ${!...} ) is used to access the array inside the function. ^[1]

By definition, this approach requires the arrays to be stored in variables up front.

As for your desire to do it without storing your arrays in a variable : You cannot pass array literals as arguments in Bash; the conceptual equivalent of that is to simply pass individual parameters .

The problem with that is that you won't be able to tell where the elements of one array end and the next one begins .

You can address that by introducing a separator parameter whose sole purpose is to indicate when an array ends.

For instance, if you know that your elements never contain a newline, you can use $\\n as the separator:

#!/usr/bin/env bash

function checkParameter(){
  local -i arrayNdx=1
  for param; do # Loop over all parameters
    [[ $param == $'\n' ]] && { (( ++arrayNdx )); continue; }
    echo "From array $arrayNdx: [$param]"
  done  
}

# Pass the array elements as individual parameters
# and use $'\n' (a newline) to separate the elements of the 1st
# array from those of the 2nd.
checkParameter "a1-child 1" "a1-child 2" "a1-child 3" $'\n' "a2-child 1" "a2-child 2"

---

^{[1] Just to show a simplified version of your solution to make the variable-indirection part clearer:}

function checkParameter(){
    for a in "${!1}"; do
        echo "${a}"
    done

    for b in "${!2}"; do
        echo "${b}"
    done
}

checkParameter a[@] b[@]