Error while passing array as function parameter in bash

Question

This is the code I am dealing with:

function execute {
    task="$1"
    servername="$2"
    "$task" "${servername[@]}" 

}


function someOtherThing {

    val=$1
    echo "$val"

}


function makeNecessaryDirectory {

    arr=("$@")
    echo "${arr[@]}"

}


dem=(1 2 3 4 5)


execute someOtherThing 1
execute makeNecessaryDirectory "${dem[@]}"

Output:

1
1

Expected output:

1
1 2 3 4 5

How to achieve this? I found no error logically.

Side question:

Is it safe to always receive 2nd parameter as an array inside execute so that it can deal with both of the dependent functions, or i should have an explicit check inside execute?

You are passing the array as individual args to execute and then only passing the first one the makeNecessaryDirectory, so $@ is just 1. — 123
– 123, Commented Sep 21, 2015 at 8:10
Wouldn't your expected output be 1, 1, 2, 3, 4, 5 (all on separate lines)? — gniourf_gniourf
– gniourf_gniourf, Commented Sep 21, 2015 at 8:45
@gniourf_gniourf, the code portion is edited on makenecessaryDirectory. — Ahasanul Haque
– Ahasanul Haque, Commented Sep 21, 2015 at 8:49
Ok. Then why don't you just use execute() { "$@"; }? it seems that's what you're after… — gniourf_gniourf
– gniourf_gniourf, Commented Sep 21, 2015 at 8:50

gniourf_gniourf · Accepted Answer · 2015-09-21 08:52:48Z

1

As explained in my comment

You are passing the array as individual args to execute and then only passing the first one the makeNecessaryDirectory, so $@ is just the single argument passed which is 1.

I would do it this way, I have added comments to the parts i have changed. It is only minor changes but should hopefully work for you.

#!/bin/bash

function execute {
    task="$1"
    servername="$2"
    "$task" "$servername"
    #No longer pass array here just pass servername as the name of array

}


function someOtherThing {

    val=$1
    echo "$val"

}


function makeNecessaryDirectory {

    local arr=("${!1}") 
    #Indirect reference to the first arg passed to function which is now the
    #name of the array

    for i in "${arr[@]}";
       do
           echo "$i"
       done

}


dem=(1 2 3 4 5)


execute someOtherThing 1
execute makeNecessaryDirectory 'dem[@]' #Pass the array name instead of it's contents

edited Sep 21, 2015 at 8:52

gniourf_gniourf

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answered Sep 21, 2015 at 8:17

123

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2 Comments

Ahasanul Haque Over a year ago

Thanks for the solution, i will use it. But how can I achieve the expected output with passing the array elements?

123 Over a year ago

@AhsanulHaque You can't really, you can't pass array elements, whilst still maintaining the array. Any embedded spaces would be lost. This is the only way i know of to preserve the array as it is between functions. You could use shift in execute, and then set servername to $@, but that is not a very good way to do it.

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