Why php does not complain when referencing a non existing variable?

Question

I was wondering why php does not complain when we reference a non existing variable (being it a plain variable or array), is this just the way it is, or there is something else I am missing? For example this code

<?php
$t = &$r["er"];
var_dump($r);
?>

throws no warning about a non existing variable.

Besides that the var_dump show this:

array(1) { ["er"]=> &NULL }

that &NULL is something I didn't really expected, I thought I would get a plain NULL.

Thanks in advance!

Answer 1

If memory of the PHP interpreter reference var allocation serves me right, PHP will create a null element in the hash table with a key like the one you sent and reference it. This is visible by running the following test:

<?php
$i = 0;
$arr = [];
$arrS = null;
$v = memory_get_peak_usage();
for ($i = 0; $i < 150; $i++) {
    $arrS = &$arr[rand()];
}
$v = memory_get_peak_usage() - $v;
echo $v;

Until the default heap size, PHP will return exactly an extra 0 memory used - as it is still allocating already "prepared" array items (PHP keeps a few extra hash table elements empty but allocated for performance purposes). You can see this by setting it from 0 to 16 (which is the heap size!).

When you get over 16, PHP will have to allocate extra items, and will do so on i=17, i=18 etc..., creating null items in order to reference them.

P.S: contrary to what people said, this does NOT throw an error, warning or notice. Recalling an empty item without reference would - referencing an inexistent item does not. Big big BIG difference.

Answer 2

throws no warning about a non existing variable.

This is how references work. $a = &$b; creates $b if it does not exist yet, "for future reference", if you will. It's the same as in parameters that are passed by reference. Maybe this looks familiar to you:

preg_match($pattern, $string, $matches);

Here the third parameter is a reference parameter, thus $matches does not have to exist at time of the method invocation, but it will be created.

that &NULL is something I didn't really expected, I thought I would get a plain NULL.

Why didn't you expect it? $r['er'] is a reference "to/from" $t. Note that references are not pointers, $r['er'] and $t are equal references to the same value, there is no direction from one to the other (hence the quotation marks in the last sentence).