Question

<?php
error_reporting(-1);
$a = array('key' => '2');
$a = $a['key'];
var_dump($a);
echo $a['key'], "\n";
?>

I expected the code to throw empty string or a warning saying key isn’t present, but the code prints 2;

PHP 5.3.10-1

Update1 : Adding var_dump and error_reporting.

Output is

hari@hari-laptop:~$ php run.php  
string(1) "2"  
2  

Answer 1

$a is a string that holds the value "2".

Strings in PHP can be indexed to access their characters using square brackets. For example:

$a = "abcde";
echo $a[3];  // This will print 'd'

When the index is a string, PHP tries to parse it as an integer. If the string is not a number, parsing will give 0. Hence, The zero’th index, representing the first character is used. (In PHP 5.4 and up, it gives a warning). In your case, you got “2” because $a['any_stuff'] = first character = '2'

From PHP manual:

String access and modification by character

Characters within strings may be accessed and modified by specifying
the zero-based offset of the desired character after the string using
square array brackets, as in $str[42]. Think of a string as an array
of characters for this purpose. The functions substr() and
substr_replace() can be used when you want to extract or replace more
than 1 character.

As of PHP 5.4 string offsets have to either be integers or integer-like strings, otherwise a warning will be thrown.