PHP :: Bug #15025 :: Copy of array is affected by reference

Bug #15025

Copy of array is affected by reference

Submitted:

2002年01月13日 22:18 UTC

Modified:

2002年10月27日 11:59 UTC

Votes:	25
Avg. Score:	4.3 ± 1.0
Reproduced:	20 of 20 (100.0%)
Same Version:	4 (20.0%)
Same OS:	5 (25.0%)

From:

jimmy at harlindong dot com

Assigned:

Status:

Wont fix

Package:

Scripting Engine problem

PHP Version:

4.3.0-dev

OS:

FreeBSD, Linux

Private report:

CVE-ID:

None

View Developer Edit

[2002年01月13日 22:18 UTC] jimmy at harlindong dot com

You can guess / see from this simple script:
$a = array(5);
$b =& $a[0];
$c = $a;
echo $c[0];
$b = 1;
echo $c[0];

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[2002年01月13日 22:32 UTC] jimmy at harlindong dot com

$c should not have been affected by the assignment to $b because $c should only be a copy to $a.
What happened is the first echo shows 5 and after assignment to $b, the content of $c changes to 1 when it should stay as 5.

[2002年09月17日 13:31 UTC] kylam at superapple dot org

Verified with 4.2.3 on Windows 2000 too.

[2002年09月17日 20:50 UTC] iliaa@php.net

updating version.

[2002年10月01日 05:47 UTC] zeev@php.net

This has to do with the fact we don't deep-search when we copy values. $a is not a reference, but has an element inside it which is a reference. Therefore, when $a is copied to $c, and no deep search is made, $c ends up having that reference inside of it as well.
It may be possible to fix, but right now I'm just wanted to add the analysis here...