<?php
ini_set('display_errors',1);
ini_set('error_reporting',-1);
$data = null;
var_export( $data['name']);
echo PHP_EOL;
var_dump($data['name']);
Why the result is null, but no notice or warning occured?
Because null is an undefined type, $data['name'] therefore creates its own array.
Just reread documentation about type casting in php.
http://php.net/manual/en/language.types.type-juggling.php
PHP is not so strict as c/c++ or java :-)
If you check the data type of $data before assigning null, you will get the undefined variable notice.
echo gettype($data);
$data = null;
After you assigned null to $data, you will see NULL for its value and its data type.
$data = null;
echo gettype($data);
var_dump($data);
According to the documentation of NULL, you're getting NULL for $data['name'] means that it has not been set to any value yet.
The special NULL value represents a variable with no value. NULL is the only possible value of type null.
A variable is considered to be null if:
it has been assigned the constant NULL.
it has not been set to any value yet.
it has been unset().
The following two examples show that the previously defined variable is not auto-converting to array, and keep its original data type.
$data = null; // $data is null
var_dump($data['name']); // null
var_dump($data[0]); // null
$data = 'far'; // $data is string
$data[0] = 'b'; // $data is still string
echo $data; // bar