Strange behavior of foreach when using reference: foreach ($a as &$v) { ... }

Question

<?php
  $a = array('a', 'b', 'c', 'd');

  foreach ($a as &$v) { }
  foreach ($a as $v) { }

  print_r($a);
?>

I think it's a normal program but this is the output I am getting:

Array
(
    [0] => a
    [1] => b
    [2] => c
    [3] => c
)

Can someone please explain this to me?

What does *"i think, its normal prog but web i saw op.........then ....ahh wat is it"* mean? — Bart Kiers, Feb 11 '11 at 13:01
possible duplicate of [Unintentional destruction of php array element?](http://stackoverflow.com/questions/4966352/unintentional-destruction-of-php-array-element) — alexn, Feb 11 '11 at 13:25
This is the best explanation for this behavior: http://schlueters.de/blog/archives/141-References-and-foreach.html — Stefan Gehrig, Feb 11 '11 at 13:26
possible duplicate of [Strange behaviour after loop by reference - Is this a PHP bug?](http://stackoverflow.com/questions/8220399/strange-behaviour-after-loop-by-reference-is-this-a-php-bug) — John Carter, May 05 '12 at 11:09
This question is now [discussed on Meta](https://meta.stackoverflow.com/q/360250/3151675). — Tamás Sengel, Dec 05 '17 at 18:40
The PHP manual recommends calling `unset($v);` after the first foreach loop to avoid this problem. http://php.net/manual/en/control-structures.foreach.php — Sean the Bean, Oct 12 '18 at 14:10

score 115 · Accepted Answer · edited Jun 20 '20 at 09:12

This is well-documented PHP behaviour See the warning on the foreach page of php.net

Warning

Reference of a $value and the last array element remain even after the foreach loop. It is recommended to destroy it by unset().

$a = array('a', 'b', 'c', 'd');

foreach ($a as &$v) { }
unset($v);
foreach ($a as $v) { }

print_r($a);

EDIT

Attempt at a step-by-step guide to what is actually happening here

$a = array('a', 'b', 'c', 'd');
foreach ($a as &$v) { }   // 1st iteration $v is a reference to $a[0] ('a')
foreach ($a as &$v) { }   // 2nd iteration $v is a reference to $a[1] ('b')
foreach ($a as &$v) { }   // 3rd iteration $v is a reference to $a[2] ('c')
foreach ($a as &$v) { }   // 4th iteration $v is a reference to $a[3] ('d')

                          // At the end of the foreach loop,
                          //    $v is still a reference to $a[3] ('d')

foreach ($a as $v) { }    // 1st iteration $v (still a reference to $a[3]) 
                          //    is set to a value of $a[0] ('a').
                          //    Because it is a reference to $a[3], 
                          //    it sets $a[3] to 'a'.
foreach ($a as $v) { }    // 2nd iteration $v (still a reference to $a[3]) 
                          //    is set to a value of $a[1] ('b').
                          //    Because it is a reference to $a[3], 
                          //    it sets $a[3] to 'b'.
foreach ($a as $v) { }    // 3rd iteration $v (still a reference to $a[3]) 
                          //    is set to a value of $a[2] ('c').
                          //    Because it is a reference to $a[3], 
                          //    it sets $a[3] to 'c'.
foreach ($a as $v) { }    // 4th iteration $v (still a reference to $a[3]) 
                          //    is set to a value of $a[3] ('c' since 
                          //       the last iteration).
                          //    Because it is a reference to $a[3], 
                          //    it sets $a[3] to 'c'.

@Manish Trivedi: See the **Warning** part for why this is happening. There's nothing wrong with your program. — BoltClock, Feb 11 '11 at 13:05
Very good explanation! Thank you :D I have to remember that one as a reference/dupe, all other answers about reference with foreach doesn't explain it as good as you did here! — Rizier123, Mar 27 '15 at 15:45
In your `unset($v);`, why is `$v` accessible outside the scope of foreachs? :o — nawfal, Nov 20 '15 at 13:15
@nawfal - don't understand the question.... PHP local variable scope is within a function, not simply within a loop — Mark Baker, Nov 20 '15 at 13:17
@MarkBaker thank you. You answered my question. Just coming to grips with PHP idiosyncrasies :) — nawfal, Nov 20 '15 at 13:40
Great answer! Glad you mentioned the workaround of adding `unset($v);` after the first foreach loop. — Sean the Bean, Oct 12 '18 at 13:59

score 3 · Answer 2 · edited Jul 03 '15 at 10:06

The first foreach loop does not make any change to the array, just as we would expect. However, it does cause $v to be assigned a reference to each of $a’s elements, so that, by the time the first loop is over, $v is, in fact, a reference to $a[2].

As soon as the second loop starts, $v is now assigned the value of each element. However, $v is already a reference to $a[2]; therefore, any value assigned to it will be copied automatically into the last element of the array!

Thus, during the first iteration, $a[2] will become zero, then one, and then one again, being effectively copied on to itself. To solve this problem, you should always unset the variables you use in your by-reference foreach loops—or, better yet, avoid using the former altogether.

Strange behavior of foreach when using reference: foreach ($a as &$v) { ... }

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