Pointers in PHP issue

Question

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Strange behavior Of foreach
Strange behaviour after loop by reference - Is this a PHP bug?

Can anybody explain me why this code:

<pre>
<?php

$a = array('page', 'email', 'comment');
$b = array('page' => 'realpage', 'email' => 'reaLmail', 'comment' => 'c');
$c = array();

foreach ($a as &$item) {
        if (isset($b[$item])) {
               $item =  $b[$item];
        }
}


foreach ($a as $item) {
        $c[] = $item;
}

print_r($c);

Outputs

 Array
(
    [0] => realpage
    [1] => reaLmail
    [2] => reaLmail
)

??? Why BEFORE second loop a is (by var_dump)

array(3) {
  [0]=>
  string(8) "realpage"
  [1]=>
  string(8) "reaLmail"
  [2]=>
  &string(1) "c"
}

But in first iteration, a is

 array(3) {
  [0]=>
  string(8) "realpage"
  [1]=>
  string(8) "reaLmail"
  [2]=>
  &string(8) "realpage"
}

and on second and third [1] and [2] indexes are the same "reaLmail", and [2] is pointer? Thank you!

BTW, there are no pointers in PHP (in userland at least), `&` is the reference operator (they're similar in C++, but not the same thing). — John Carter, May 05 '12 at 11:08
if you place `var_dump($a);` after first `foreach` you'll see that last item of array is actually a poiner. So last ` $c[] = $item;` doesn't work and `$item` remains the same from previous iteration. — s.webbandit, May 05 '12 at 11:08

score 1 · Accepted Answer · answered May 05 '12 at 11:08

1

if you use foreach (... as &..), then unset is required as describe in the php manual:

foreach ($a as &$item) {
        if (isset($b[$item])) {
               $item =  $b[$item];
        }
}
unset($item);

answered May 05 '12 at 11:08

meze

14,975
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score 0 · Answer 2 · answered May 05 '12 at 11:05

0

You don't need the & in the first loop, which makes it a reference. Remove it, and your example should work fine.

foreach ($a as &$item)

answered May 05 '12 at 11:05

Nadh

6,987
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Pointers in PHP issue

2 Answers2