PHP: check if object/array is a reference

Question

Sorry to ask, its late and I can't figure a way to do it... anyone can help?

$users = array(
    array(
        "name" => "John",
        "age"   => "20"
    ),
    array(
        "name" => "Betty",
        "age"   => "22"
    )
);

$room = array(
    "furniture" => array("table","bed","chair"),
    "objects"   => array("tv","radio","book","lamp"),
    "users" => &$users
);

var_dump $room shows:

...
'users' => &
...

Which means "users" is a reference.

I would like to do something like this:

foreach($room as $key => $val) {
    if(is_reference($val)) unset($room[$key]);
}

The main goal is to copy the array WITHOUT any references.

Is that possible?

Thank you.

Answer 1

Reflection API's isPassedByReference might be interested here. For example, here is how I used it in a project:

$refMethod = new ReflectionMethod($class_name,  '__construct');
$params = $refMethod->getParameters();

$re_args = array();

foreach($params as $key => $param)
{
    if ($param->isPassedByReference())
    {
        // it is passed by reference
    }
    else
    {
        // it is not...
    }
}

Answer 2

You can test for references in a multi-dimensional array by making a copy of the array, and then altering and testing each entry in turn:

$roomCopy = $room;
foreach ($room as $key => $val) {
  $roomCopy[$key]['_test'] = true;
  if (isset($room[$key]['_test'])) {
    // It's a reference
    unset($room[$key]);
  }
}
unset($roomCopy);

With your example data, $room['furniture'] and $roomCopy['furniture'] will be separate arrays (as $roomCopy is a copy of $room), so adding a new key to one won't affect the other. But, $room['users'] and $roomCopy['users'] will be references to the same $users array (as it's the reference that's copied, not the array), so when we add a key to $roomCopy['users'] it is visible in $room['users'].

Answer 3

something recursive maybe.

function removeReferences($inbound)
{
    foreach($inbound as $key => $context)
    {
        if(is_array($context))
        {
            $inbound[$key] = removeReferences($context)
        }elseif(is_object($context) && is_reference($context))
        {
            unset($inbound[$key]); //Remove the entity from the array.
        }
    }
    return $inbound;
}

Answer 4

The best I can manage is a test of two variables to determine if one is a reference to the other:

$x = "something";
$y = &$x;
$z = "something else";

function testReference(&$xVal,&$yVal) {
    $temp = $xVal;
    $xVal = "I am a reference";
    if ($yVal == "I am a reference")  { echo "is reference<br />"; }  else  { echo "is not reference<br />"; }
    $xVal = $temp;
}

testReference($x,$y);
testReference($y,$x);

testReference($x,$z);
testReference($z,$x);

testReference($y,$z);
testReference($z,$y);

but I doubt if it's much help

Really dirty method (not well tested either):

$x = "something";
$y = &$x;
$z = "something else";

function isReference(&$xVal) {
    ob_start();
    debug_zval_dump(&$xVal);
    $dump = ob_get_clean();
    preg_match('/refcount\((\d*)\)/',$dump,$matches);
    if ($matches[1] > 4) { return true; } else { return false; }
}

var_dump(isReference($x));
var_dump(isReference($y));
var_dump(isReference($z));

To use this last method in your code, you'd need to do something like:

foreach($room as $key => $val) {
    if(isReference($room[$key])) unset($room[$key]);
}

because $val is never a reference as it's a copy of the original array element; and using &$val makes it always a reference