Question

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.

Answer 1

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" ;-))

Answer 2

Here s an explanation from the man himself:

$y = "some test";

foreach ($myarray as $y) {
    print "$y
";
}
Here $y is a symbol table entry referencing a string containing "some test". On the first iteration you essentially do:
$y = $myarray[0];  // Not necessarily 0, just the 1st element
So now the storage associated with $y is overwritten by the value from $myarray. If $y is associated with some other storage through a reference, that storage will be changed.

Now let s say you do this:
$myarray = array("Test");
$a = "A string";
$y = &$a;

foreach ($myarray as $y) {
    print "$y
";
}
Here $y is associated with the same storage as $a through a reference so when the first iteration does:
$y = $myarray[0];
The only place that "Test" string can go is into the storage associated with $y.

Answer 3

This is how you would fix this problem:

foreach($array as &$element)
{
    $element *= 2;
}
unset($element); #gets rid of the reference and cleans the var for re-use.

foreach($array as $element) { }

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