Bug #68914 [Opn->Nab]: Reference survives beyond function scope
| From: | requinix@php.net | Date: | Mon, 26 Jan 2015 07:00:27 +0000 |
| Subject: | Bug #68914 [Opn->Nab]: Reference survives beyond function scope | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-190224@lists.php.net to get a copy of this message | ||
Edit report at https://bugs.php.net/bug.php?id=68914&edit=1
ID: 68914
Updated by: requinix@php.net
Reported by: steve dot piner at signify dot co dot nz
Summary: Reference survives beyond function scope
-Status: Open
+Status: Not a bug
Type: Bug
-Package: PHP Language Specification
+Package: Scripting Engine problem
Operating System: Ubuntu 14.10
PHP Version: 5.6.5
Block user comment: N
Private report: N
New Comment:
It survives because $x is static and each member in $x is a reference.
$x was initialized once to that range(1,3) array. Whenever f() returns it will return a copy of that
array, thus $z and $y have both the same array values even though they are different actual arrays.
Modifying $y would have no effect on $z because of the copying... if it weren't for the by-ref
foreach you did, which caused each member in the array to become a reference. That makes $x[2],
$y[2], and $z[2] all the same so a change made to one of them will "appear" in the others.
Previous Comments:
------------------------------------------------------------------------
[2015-01-26 05:23:23] steve dot piner at signify dot co dot nz
Description:
------------
A reference created with a foreach loop appears to survive long after the function creating the
reference has exited.
Test script:
---------------
function f() {
static $x;
if(!$x) {
$x = range(1, 3);
}
foreach($x as $i => &$j) { }
return $x;
}
$z = f();
$y = f();
$y[2] = 'surprise';
print $z[2] . "\n";
Expected result:
----------------
3
Actual result:
--------------
surprise
------------------------------------------------------------------------
--
Edit this bug report at https://bugs.php.net/bug.php?id=68914&edit=1