display array in a function - not working

Question

What am I missing here?

I have created a simple array:

declare -a appArray=(
    "item1 -a -b"
    "item2 -c -d"
    )

If I echo this I can see it all

echo ${appArray[@]}

> item1 -a -b item2 -c -d

I then create a function as follows:

fc_DEBUG () { 
    if [ $1 -eq 1 ] ; then 
        echo $2; 
    fi; 
};

It is designed to sit in a bash script, and if I set a DEBUG variable it will echo the text back. So I can use it throughout the script without needing to manually add / remove things.

It works fine with basic data: e.g

fc_DEBUG $DEBUG "This is DEBUG text"

If I call this with the Array however, I only get a part of the Array.

fc_DEBUG $DEBUG "${appArray[@]}"

> item1 -a -b

Answer 1

${appArray[@]} gets expanded before fc_DEBUG runs. So the second argument the function sees, is the first of the array. To be explicit, the three arguments fc_DEBUG sees, are

$DEBUG "item1 -a -b" "item2 -c -d"

(replace $DEBUG with the words resulting from the split+glob operator applied to the actual value of $DEBUG (as you forgot to quote it)). In technical terms, the array is passed by value, not by reference.

fc_DEBUG () { 
    if [ "$1" -eq 1 ] ; then 
        shift
        echo "$@"
    fi
}

Now, the first argument is dropped from the argument list with shift, and the rest of all the arguments is printed.

Call it with a quoted array:

fc_DEBUG "$DEBUG" "${appArray[@]}"