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Bug #17997 Warning when switch & reference are combined
Submitted: 2002-06-26 11:25 UTC Modified: 2003-10-07 20:51 UTC
Votes:2
Avg. Score:3.0 ± 2.0
Reproduced:1 of 2 (50.0%)
Same Version:1 (100.0%)
Same OS:1 (100.0%)
From: Xuefer at 21cn dot com Assigned:
Status: Closed Package: Scripting Engine problem
PHP Version: 4CVS, 5CVS OS: All
Private report: No CVE-ID: None
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From: Xuefer at 21cn dot com
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 [2002-06-26 11:25 UTC] Xuefer at 21cn dot com
let the sample code say the problem
<?
error_reporting(E_ALL);

$str1 = '.........$...';
$str = &$str1;
$l = strlen($str);
for ($i = 0; $i < $l; $i ++)
{
	switch ($str{$i})
	{
	case '$':
		$str = '123'; // change string
		$i = -1; // set to -1, for next loop to be 0
		$l = strlen($str); // u know what this do
		break;

	default:
		echo $str{$i};
	}
}
?>

-----------
result:
-----------
.........
Notice: Uninitialized string offset: 9 in d:\myfile\wwwroot\our-sky.com\test.php on line 19
123


how ever, if i change
	$str = '123';
to:
	$s = '123';
	$str = &$s;

no wanning

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 [2002-07-23 16:45 UTC] hholzgra@php.net
the error message is on the closing '}' of the switch ...
does switch re-evaluate its argument on termination?
this is scary!
 [2003-08-10 18:12 UTC] helly@php.net
Yes it refetches the variable passed to switch() to check whether it must be freed or not.
 [2003-10-07 12:01 UTC] zeev@php.net
This bug has been fixed in CVS.

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Fixed
 [2003-10-07 20:51 UTC] sniper@php.net
Fixed in PHP 5, won't fix in PHP 4.

 
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