php bug?

[Chameleon]

In the code below I merge array $b with each element of array $a.
So array $a must be after:
$a == array(array(1, 2, 6, 7, 8), array(2, 3, 6, 7, 8), array(4, 5, 6,
7, 8));

instead, it seems to overwrite last element of $a with before-last and
we have:
$a == array(array(1, 2, 6, 7, 8), array(2, 3, 6, 7, 8), array(2, 3, 6,
7, 8));

php version: PHP 5.1.0RC1 (cli) (built: Aug 16 2005 13:51:15)
(the windows version)

It is a php bug or a my bug?

I see that var_dump($a); prints the corrent data but when I print
element-element var_dump it prints the before-last element two times and
last element zero times.

----------------------------------------------
<?
$a = array(array(1, 2), array(2, 3), array(4, 5));
$b = array(6, 7, 8);

foreach($a as & $v)
$v = array_merge($v, $b);

var_dump($a);
foreach($a as $v)
var_dump($v);
?>
----------------------------------------------

[Janwillem Borleffs]

Chameleon wrote:

I see that var_dump($a); prints the corrent data but when I print
element-element var_dump it prints the before-last element two times
and last element zero times.

Unreproducable with version 5.1.1; upgrade to this version first, then try
it again.
JW

[Toby Inkster]

Chameleon wrote:

$a == array(

^^

Should be a single equals sign, right?

--
Toby A Inkster BSc (Hons) ARCS
Contact Me ~ http://tobyinkster.co.uk/contact

[Chameleon]

Toby Inkster wrote:

Chameleon wrote:

$a == array(

^^

Should be a single equals sign, right?

So array $a must be after:
$a == array(array(1, 2, 6, 7, 8), array(2, 3, 6, 7, 8), array(4, 5, 6,
7, 8));
I mean:
So array $a must be after equals to array(array(1, 2, 6, 7, 8), array(2,
3, 6, 7, 8), array(4, 5, 6, 7, 8));

[Chung Leong]

Chameleon wrote:

It is a php bug or a my bug?

It's your bug I suspect. The problem can't be reproduced with the code
as given. If you move the var_dump($a) to the end, then you see it.

<?

$a = array(array(1, 2), array(2, 3), array(4, 5));
$b = array(6, 7, 8);

foreach($a as & $v)
$v = array_merge($v, $b);

foreach($a as $v)
var_dump($v);

var_dump($a);

?>

The second foreach() is setting the value of $v to each element in $a.
At this point, $v is still pointing to the last element in $a,
resulting from the first foreach(). The last element in $a is thus is
set repeatedly to all elements preceding it.

It's a good idea to unset() references when they're no longer needed to
avoid unintended modification later on.

[Mara Guida]

Chameleon wrote:

----------------------------------------------
<?
$a = array(array(1, 2), array(2, 3), array(4, 5));
$b = array(6, 7, 8);

foreach($a as & $v)
$v = array_merge($v, $b);
/* references or no references I don't like to change
* the values that control the foreach() loop.
* I changed your foreach() code and it worked here.
*/
$tmp = array();
foreach($a as $v)
$tmp[] = array_merge($v, $b);
$a = $tmp;
unset($tmp);

var_dump($a);
foreach($a as $v)
var_dump($v);
?>
----------------------------------------------