php undefined index notice not raised when indexing null variable

Question

I'm curious to know if the following behaviour in PHP is intended or not. And, if it is intended, it is considered acceptable to initialize an array from a null variable by creating an index into it (as is done in the first code snippet)?

error_reporting(E_ALL);
$arr = null;

echo ($arr["blah"]===null) ? "null" : $arr["blah"];

$arr["blah"] = "somevalue";
echo "<br>";
echo ($arr["blah"]===null) ? "null" : $arr["blah"];
var_dump ($arr);

This outputs

null
somevalue

array (size=1)
   'blah' => string 'somevalue' (length=9)

However, if the array is initialized first (see code below), I get the exact same output, but an "Undefined Index" notice is given when I first try $arr["blah"]

error_reporting(E_ALL);
$arr = array();

echo ($arr["blah"]===null) ? "null" : $arr["blah"];

$arr["blah"] = "somevalue";
echo "<br>";
echo ($arr["blah"]===null) ? "null" : $arr["blah"];
var_dump ($arr);

Answer 1

PHP won't attempt the comparison if the array is null.

In the second circumstance, a comparison does occur because the array is set. PHP does not check to see if it is empty.

Your ternary is attempting to access the variable $arr["blah"], not checking to see if it is set before doing a comparison.

The proper way to write this would be:

error_reporting(E_ALL);
$arr = array();

if(isset($arr["blah"])) echo ($arr["blah"]===null) ? "null" : $arr["blah"];

$arr["blah"] = "somevalue";
echo "<br>";
if(isset($arr["blah"])) echo ($arr["blah"]===null) ? "null" : $arr["blah"];
var_dump ($arr);

Answer 2

Actually, John Vargo was correct. If a variable is null, accessing it as if it were an array will simply return null without notices. This will change in the upcoming 7.4 version, then it will produce a notice.

Notice: Trying to access array offset on value of type null

The actual output is still the same.