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Bug #49068 Using undefined assoc array index does not cause notice inside foreach
Submitted: 2009-07-26 22:09 UTC Modified: 2009-07-27 10:24 UTC
From: simon+php at thulbourn dot com Assigned:
Status: Not a bug Package: Scripting Engine problem
PHP Version: 5.*, 6SVN (2009-07-26) OS: *
Private report: No CVE-ID: None
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 [2009-07-26 22:09 UTC] simon+php at thulbourn dot com 
Description:
------------
An array with no associative keys doesn't throw a notice or error when 
one is used as an associative keyed array.

Reproduce code:
---------------
<?php

	$array = array('foo', 'bar', 'baz');
	
	foreach ($array as $a) {
		echo $a['foo'];
	}

?>

Expected result:
----------------
Notice: Undefined index

Actual result:
--------------
fbb

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 [2009-07-26 23:07 UTC] rasmus@php.net 
That has nothing to do with a foreach.  The same effect can be seen with:

$a = 'foo';
echo $a['bar'];

$a is a string which is essentially an array of characters with numerical indices.  Given that, 'bar' is converted to an integer, so you end up with $a[0] which does exist, so no warning.
 [2009-07-27 10:04 UTC] simon+php at thulbourn dot com 
I believe that it should issue a warning rather than converting to an 
integer.

It seems like a retarded action to convert the string to a 0, if I was 
to use a numeric index that doesn't exist, I'd receive a notice. Some 
consistency would be nice.)
 [2009-07-27 10:20 UTC] simon+php at thulbourn dot com 
Granted, I could have provided a better sample case so:

$a = array('foo');

// echo a index that doesn't exist
echo $a['bar'];

PHP should produce a notice claiming an undefined index 'bar', instead 
it outputs the first char of the 0 index, in this case 'f', it's 
obviously converting the string to a 0 instead of producing the notice.)
 [2009-07-27 10:24 UTC] simon+php at thulbourn dot com 
Sorry, the new test code was wrong, here's one that proves my point

$blah = "foo";echo $blah["frog"];)
 
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