Edit report at http://bugs.php.net/bug.php?id=54321&edit=1
ID: 54321
Updated by: il...@php.net
Reported by: unokpasabaxaki at yahoo dot es
Summary: Foreach not running through added elements when
passed by reference
-Status: Open
+Status: Bogus
Type: Bug
Package: *Programming Data Structures
PHP Version: 5.2.17
Block user comment: N
Private report: N
New Comment:
Thank you for taking the time to write to us, but this is not
a bug. Please double-check the documentation available at
http://www.php.net/manual/ and the instructions on how to report
a bug at http://bugs.php.net/how-to-report.php
foreach makes a copy of the array to run through, any changes to that
array within
the scope of foreach() construct are not visible to it.
Previous Comments:
------------------------------------------------------------------------
[2011-03-20 13:45:06] unokpasabaxaki at yahoo dot es
Description:
------------
I found out that foreach sometimes doesn't run through elements added to
an array
when the value is passed by reference. I haven't tested enough to say it
for sure,
but I think it happens only on elements added on what would be the last
iteration
of foreach.
Test script:
---------------
$array = array('a', 'b', 'c');
foreach ($array as $i => &$val) {
echo $val.' ';
if ($i % 2 == 1) $array[] = $val.'2';
}
Expected result:
----------------
When it reaches $array[1] (b), it creates $array[3] with the value "b2".
Then it
runs through $array[3] and creates $array[4] with the value "b22". After
that, it
should run through $array[4] too, outputting altogether "a b c b2 b22"
and
stopping.
Actual result:
--------------
Since at the start of the iteration, $array[3] is the last element of
the array,
after $array[4] is created in that iteration, it doesn't run through it,
outputting just "a b c b2". Still, $array[4] is created.
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/bug.php?id=54321&edit=1
