PHP 5.4 ForEach madness?

Question

I'm not sure how to word it.

Take the following example.

<?
$arr = array('first','second','third', 'fourth');
foreach ($arr as $k=>&$n) {
    $n = $n;
}
var_dump($arr);
>?

returns

array(4) {
  [0]=>
  string(5) "first"
  [1]=>
  string(6) "second"
  [2]=>
  string(5) "third"
  [3]=>
  &string(6) "fourth"
}

That makes total sense, why would anything change? Well, adding a second loop that doesn't even do anything is where I got lost.

<?
$arr = array('first','second','third', 'fourth');
foreach ($arr as $k=>&$n) {
    $n = $n;
}
foreach ($arr as $k=>$n) {
;
}
var_dump($arr);
?>

returns

array(4) {
  [0]=>
  string(5) "first"
  [1]=>
  string(6) "second"
  [2]=>
  string(5) "third"
  [3]=>
  &string(5) "third"
}

The last value of the array has become the same as the second to last. Why?

http://stackoverflow.com/questions/4969243/strange-behavior-of-foreach/4969286#4969286 — Mark Baker, Mar 19 '14 at 14:39
[Here is a nice description of exactly this effect](http://schlueters.de/blog/archives/141-References-and-foreach.html) and why it happens with reference iteration. — Digital Chris, Mar 19 '14 at 14:46

score 6 · Accepted Answer · answered Mar 19 '14 at 14:39

6

You're turning $n into a reference:

foreach ($arr as $k=>&$n) {
                     ^----

Once a PHP variable is "referenced" like that, it STAYS a reference, so in your next loop:

foreach ($arr as $k=>$n) {

That $n will still be a reference to the LAST item you iterated over in the first loop.

answered Mar 19 '14 at 14:39

Marc B

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That's a great way to word it! Thanks. – Mar 19 '14 at 14:40

score 1 · Answer 2 · edited May 23 '17 at 11:44

1

You got the address operator & before $n in the first loop. Removing that will solve the problem.

This problem is explained here:

Why php iteration by reference returns a duplicate last record?

edited May 23 '17 at 11:44

Community

1
1

answered Mar 19 '14 at 14:38

thatidiotguy

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@Jhawins What do you mean, that is the exact problem right now. I just ran the code without the & and it works as expected. – thatidiotguy Mar 19 '14 at 14:39
I'm in the middle of something. I don't see how that could be it but I'll read it – Mar 19 '14 at 14:39
I get it now. That's odd. Thanks! – Mar 19 '14 at 14:42

score 0 · Answer 3 · answered Mar 19 '14 at 14:37

A stab in the dark but maybe try resetting the internal pointer of the array using reset() ?

http://www.php.net/reset

<?
$arr = array('first','second','third', 'fourth');
foreach ($arr as $k=>&$n) {
    $n = $n;
}

reset($arr);
foreach ($arr as $k=>$n) {
;
}

var_dump($arr);
?>

PHP 5.4 ForEach madness?

3 Answers3