Question

foreach((array)$arr['subarr'] as &$foo)
 ....

…doesn’t not work. It throws a parse error.

Why?

Answer 1

You cannot cast to array and at the same time use the items as a reference.

What happens to $foo it it really isn’t an array?

The casting only applies to the loop.

$arr['subarr'] = array('one', 'two');

// make sure we have an array
if (!is_array($arr['subarr'])) {
    $arr['subarr'] = array($arr['subarr']);
}

foreach($arr['subarr'] as &$foo) {
    print($foo);
}