Passing arrays as parameters in bash

Solution 1:

You can pass multiple arrays as arguments using something like this:

takes_ary_as_arg()
{
    declare -a argAry1=("${!1}")
    echo "${argAry1[@]}"

    declare -a argAry2=("${!2}")
    echo "${argAry2[@]}"
}
try_with_local_arys()
{
    # array variables could have local scope
    local descTable=(
        "sli4-iread"
        "sli4-iwrite"
        "sli3-iread"
        "sli3-iwrite"
    )
    local optsTable=(
        "--msix  --iread"
        "--msix  --iwrite"
        "--msi   --iread"
        "--msi   --iwrite"
    )
    takes_ary_as_arg descTable[@] optsTable[@]
}
try_with_local_arys

will echo:

sli4-iread sli4-iwrite sli3-iread sli3-iwrite  
--msix  --iread --msix  --iwrite --msi   --iread --msi   --iwrite

Edit/notes: (from comments below)

descTable and optsTable are passed as names and are expanded in the function. Thus no $ is needed when given as parameters.
Note that this still works even with descTable etc being defined with local, because locals are visible to the functions they call.
The ! in ${!1} expands the arg 1 variable.
declare -a just makes the indexed array explicit, it is not strictly necessary.

Solution 2:

Note: This is the somewhat crude solution I posted myself, after not finding an answer here on Stack Overflow. It allows for only one array being passed, and it being the last element of the parameter list. Actually, it is not passing the array at all, but a list of its elements, which are re-assembled into an array by called_function(), but it worked for me. Somewhat later Ken posted his solution, but I kept mine here for "historic" reference.

calling_function()
{
    variable="a"
    array=( "x", "y", "z" )
    called_function "${variable}" "${array[@]}"
}

called_function()
{
    local_variable="${1}"
    shift
    local_array=("${@}")
}

Solution 3:

Commenting on Ken Bertelson solution and answering Jan Hettich:

How it works

the takes_ary_as_arg descTable[@] optsTable[@] line in try_with_local_arys() function sends:

This is actually creates a copy of the descTable and optsTable arrays which are accessible to the takes_ary_as_arg function.
takes_ary_as_arg() function receives descTable[@] and optsTable[@] as strings, that means $1 == descTable[@] and $2 == optsTable[@].
in the beginning of takes_ary_as_arg() function it uses ${!parameter} syntax, which is called indirect reference or sometimes double referenced, this means that instead of using $1's value, we use the value of the expanded value of $1, example:
```
baba=booba
variable=baba
echo ${variable} # baba
echo ${!variable} # booba
```
likewise for $2.
putting this in argAry1=("${!1}") creates argAry1 as an array (the brackets following =) with the expanded descTable[@], just like writing there argAry1=("${descTable[@]}") directly. the declare there is not required.

N.B.: It is worth mentioning that array initialization using this bracket form initializes the new array according to the IFS or Internal Field Separator which is by default tab, newline and space. in that case, since it used [@] notation each element is seen by itself as if he was quoted (contrary to [*]).

My reservation with it

In BASH, local variable scope is the current function and every child function called from it, this translates to the fact that takes_ary_as_arg() function "sees" those descTable[@] and optsTable[@] arrays, thus it is working (see above explanation).

Being that case, why not directly look at those variables themselves? It is just like writing there:

argAry1=("${descTable[@]}")

See above explanation, which just copies descTable[@] array's values according to the current IFS.

In summary

This is passing, in essence, nothing by value - as usual.

I also want to emphasize Dennis Williamson comment above: sparse arrays (arrays without all the keys defines - with "holes" in them) will not work as expected - we would loose the keys and "condense" the array.

That being said, I do see the value for generalization, functions thus can get the arrays (or copies) without knowing the names:

for ~"copies": this technique is good enough, just need to keep aware, that the indices (keys) are gone.
for real copies: we can use an eval for the keys, for example:
```
eval local keys=(\${!$1})
```

and then a loop using them to create a copy. Note: here ! is not used it's previous indirect/double evaluation, but rather in array context it returns the array indices (keys).

and, of course, if we were to pass descTable and optsTable strings (without [@]), we could use the array itself (as in by reference) with eval. for a generic function that accepts arrays.

Solution 4:

The basic problem here is that the bash developer(s) that designed/implemented arrays really screwed the pooch. They decided that ${array} was just short hand for ${array[0]}, which was a bad mistake. Especially when you consider that ${array[0]} has no meaning and evaluates to the empty string if the array type is associative.

Assigning an array takes the form array=(value1 ... valueN) where value has the syntax [subscript]=string, thereby assigning a value directly to a particular index in the array. This makes it so there can be two types of arrays, numerically indexed and hash indexed (called associative arrays in bash parlance). It also makes it so that you can create sparse numerically indexed arrays. Leaving off the [subscript]= part is short hand for a numerically indexed array, starting with the ordinal index of 0 and incrementing with each new value in the assignment statement.

Therefore, ${array} should evaluate to the entire array, indexes and all. It should evaluate to the inverse of the assignment statement. Any third year CS major should know that. In that case, this code would work exactly as you might expect it to:

declare -A foo bar
foo=${bar}

Then, passing arrays by value to functions and assigning one array to another would work as the rest of the shell syntax dictates. But because they didn't do this right, the assignment operator = doesn't work for arrays, and arrays can't be passed by value to functions or to subshells or output in general (echo ${array}) without code to chew through it all.

So, if it had been done right, then the following example would show how the usefulness of arrays in bash could be substantially better:

simple=(first=one second=2 third=3)
echo ${simple}

the resulting output should be:

(first=one second=2 third=3)

Then, arrays could use the assignment operator, and be passed by value to functions and even other shell scripts. Easily stored by outputting to a file, and easily loaded from a file into a script.

declare -A foo
read foo <file

Alas, we have been let down by an otherwise superlative bash development team.

As such, to pass an array to a function, there is really only one option, and that is to use the nameref feature:

function funky() {
    local -n ARR

    ARR=$1
    echo "indexes: ${!ARR[@]}"
    echo "values: ${ARR[@]}"
}

declare -A HASH

HASH=([foo]=bar [zoom]=fast)
funky HASH # notice that I'm just passing the word 'HASH' to the function

will result in the following output:

indexes: foo zoom
values: bar fast

Since this is passing by reference, you can also assign to the array in the function. Yes, the array being referenced has to have a global scope, but that shouldn't be too big a deal, considering that this is shell scripting. To pass an associative or sparse indexed array by value to a function requires throwing all the indexes and the values onto the argument list (not too useful if it's a large array) as single strings like this:

funky "${!array[*]}" "${array[*]}"

and then writing a bunch of code inside the function to reassemble the array.