The following program displays a notice (error_reporting is set to display all warnings):

-------------
<pre>
<?php
$a='hugp';
for ($i=0; isset($a[$i]); $i++) {
       echo "\nChar $i:$a[$i]";
}

?>
-------------

Displays the message

Notice:  Uninitialized string offset:  4 ... line 4


This is a bug, cause isset() is used to determine, if $a[4] is set or not, so there has to be no message, that a var is not set. It happens also, if you write $a{$i} instead of $a[$i] (this is the syntax, which will only be supported from PHP5 up).

Cause isset() is used in empty(), empty has also this bug.

This bug has already been reported (#16528), but has been closed - perhaps due to completly other example (array depending). I have been told that it's not a bug. But as you directly see with this example it is. I have written a mail to Derick Redhans who cleaned it, that it might be a fault to clean it, but never heard an answer.

Please fix this bug, cause programmers, who write the following:

$a='';
if ( !empty($a['bla']) )

will be complained with this Notice, even it is correct code (and maybe that, what he wants). Of course it is correct to display the notice, if you write:

echo $a['bla'];


---------
A small suggestion: Cause Zeev said on PHP-Congress, that the above syntax (with [] instead of {}) to exercise a char in a string is not supported any more in PHP5, it might be a very good idea to implement a switch, which will warn if you use the syntax right now! This switch could also be used, to warn the programmer for other incompatibilities... with PHP5.