Bash create parameter named array within function

Question

I'm attempting to write a function that writes arrays with a name that's passed in. Given the following bash function:

function writeToArray {
    local name="$1"
    echo "$name"
    declare -a "$name"
    ${name[0]}="does this work?"      
}

Running like this:

writeToArray $("test")

I get two errors:

bash: declare: `': not a valid identifier
=does this work?: command not found

I am expecting to be able to do this:

writeToArray $("test")
for item in "${test[@]}"; do
        echo "item"
        echo "$item"
done

This should print:

item
does this work?

How could I properly configure this to write the array (named test in the example, such that this array named test is readable outside the function)?

Answer 1

You'd use a nameref for that:

writeToArray() {
    local -n writeToArray_name="$1"
    writeToArray_name[0]="does this work?"      
}

Testing:

bash-5.0$ test[123]=qwe
bash-5.0$ writeToArray test
bash-5.0$ typeset -p test
declare -a test=([0]="does this work?" [123]="qwe")

With older versions of bash which didn't have namerefs yet, you could use eval:

writeToArray() {
  eval "$1[0]='does this work?'"
}

When writeToArray is invoked with test as argument, eval is passed test[0]='does this work?' as argument, and that in turns is evaluated as code in the shell language where it assigns does this work? to the element of index 0 of the test array (also works for associative array; scalar variables are converted to arrays).

Note that $("test") is syntax to capture and expand the output of test command and split+glob it when in list context. test (aka [) produces no output when not passed any argument), so $("test") expands to the empty string and split+glob gives you nothing at all out of it.

Here, it's the name of your variable that you want to pass to writeToArray, so test, not its contents ("$test") nor the output of a command by the same name ("$(test)") let alone the output of a command by the same name subject to split+glob as with your $("test") attempt.