foreach and reference

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

I do have a problem with Foreach, I can't understand the second foreach.

why the third part of $foo array has 'two' in its value!

In as much as we know, if we remove the '&'(in first foreach) the problem will solve, but why?

<?php

$foo = array('one', 'two', 'three');

foreach ($foo as &$bar)
{
// no-op
}

var_dump($foo);

foreach ($foo as $bar)
{
 // no-op
}

var_dump($foo);
?>
###############################################

produces:

 array(3) {
   [0]=>string(3) "one"
   [1]=>string(3) "two"
   [2]=>&string(5) "three"
  }
   array(3) {
  [0]=>string(3) "one"
  [1]=>string(3) "two"
  [2]=>&string(3) "two"
  }

That second foreach is basically: `foreach ($foo as $foo[2])`. So each iteration, you are updating the last element of $foo with the current element. And since the last element is lost, you ultimately end up with the second-to-last in the last position. — Matthew, Jul 25 '12 at 19:18

score 2 · Accepted Answer · answered Jul 25 '12 at 19:17

2

From the docs:

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

So in your case:

<?php

$foo = array('one', 'two', 'three');

foreach ($foo as &$bar)
{
    // no-op
}

var_dump($foo);

unset($bar);

foreach ($foo as $bar)
{
 // no-op
}

var_dump($foo);
?>

answered Jul 25 '12 at 19:17

PeeHaa

71,436
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I'd have called this a bug, didn't expect it to be documented. While testing I also discovered that `unset` in-between the loops does indeed fix the problem. – Joost Jul 25 '12 at 19:18
Thank you, I know we can use unset. But I don't understand why does the '&' change the value of array?! – Mehdi Raash Jul 25 '12 at 19:23
See [Matthew's comment](http://stackoverflow.com/questions/11656915/foreach-and-reference/11656964#comment15447102_11656915). – PeeHaa Jul 25 '12 at 19:24
so This is a bug.thanks. – Mehdi Raash Jul 26 '12 at 04:57
1

@wandad No it isn't a bug. It is expected behavior. – PeeHaa Jul 26 '12 at 16:52

score 0 · Answer 2 · edited May 23 '17 at 11:53

0

In the first foreach you're using a reference with '&' symbol, so the values are actually modified. I don't know what it inside of first foreach but you're overwriting old values, that's why in your second foreach values are not the same as in the beginning.

Edit There's a neat answer in this post, or just use unset as recommended

edited May 23 '17 at 11:53

Community

1
1

answered Jul 25 '12 at 19:16

Jorge Zapata

2,316
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1

Have you tried running the code in the question? It produces that output, and there's no (obvious) over-writing going on, just the two foreach loops. – andrewsi Jul 25 '12 at 19:19

foreach and reference

2 Answers2