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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T07:14:12+00:00 2026-05-14T07:14:12+00:00

If I iterate through an array twice, once by reference and then by value,

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If I iterate through an array twice, once by reference and then by value, PHP will overwrite the last value in the array if I use the same variable name for each loop. This is best illustrated through an example:

$array = range(1,5);
foreach($array as &$element)
{
  $element *= 2;
}
print_r($array);
foreach($array as $element) { }
print_r($array);

Output:

Array ( [0] => 2 [1] => 4 [2] => 6 [3] => 8 [4] => 10 )

Array ( [0] => 2 [1] => 4 [2] => 6 [3] => 8 [4] => 8 )

Note that I am not looking for a fix, I am looking to understand why this is happening. Also note that it does not happen if the variable names in each loop are not each called $element, so I’m guessing it has to do with $element still being in scope and a reference after the end of the first loop.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-14T07:14:13+00:00Added an answer on May 14, 2026 at 7:14 am

    After the first loop $element is still a reference to the last element/value of $array.
    You can see that when you use var_dump() instead of print_r()

    array(5) {
      [0]=>
      int(2)
    ...
      [4]=>
      &int(2)
    }
    

    Note that & in &int(2).
    With the second loop you assign values to $element. And since it’s still a reference the value in the array is changed, too. Try it with

    foreach($array as $element)
    {
      var_dump($array);
    }
    

    as the second loop and you’ll see.
    So it’s more or less the same as

    $array = range(1,5);
    $element = &$array[4];
    $element = $array[3];
    // and $element = $array[4];
    echo $array[4];
    

    (only with loops and multiplication …hey, I said “more or less” ;-))

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