PHP Pass by reference in foreach [duplicate]

Answer 1

I got here just by accident and the OP's question got my attention. Unfortunately I do not understand any of the explanations from the top. Seems to me like everybody knows it, gets it, accetps it, just cannot explain.

Luckily, a pure sentence from PHP documentation on foreach makes this completely clear:

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

Answer 2

Because if you create a reference to a variable, all names for that variable (including the original) BECOME REFERENCES.

Answer 3

This question has a lot of explanations provided, but no clear examples of how to solve the problem that this behavior causes. In most cases, you'll probably want the following code in your pass by reference foreach.

foreach ($array as &$row) {
    // Do stuff
}
// Unset to remove the reference
unset($row);

Answer 4

Because on the second loop, $v is still a reference to the last array item, so it's overwritten each time.

You can see it like that:

$a = array ('zero','one','two', 'three');

foreach ($a as &$v) {

}

foreach ($a as $v) {
  echo $v.'-'.$a[3].PHP_EOL;
}

As you can see, the last array item takes the current loop value: 'zero', 'one', 'two', and then it's just 'two'... : )

Answer 5

I think this code show the procedure more clear.

<?php

$a = array ('zero','one','two', 'three');

foreach ($a as &$v) {
}

var_dump($a);

foreach ($a as $v) {
  var_dump($a);
}

Result: (Take attention on the last two array)

array(4) {
  [0]=>
  string(4) "zero"
  [1]=>
  string(3) "one"
  [2]=>
  string(3) "two"
  [3]=>
  &string(5) "three"
}
array(4) {
  [0]=>
  string(4) "zero"
  [1]=>
  string(3) "one"
  [2]=>
  string(3) "two"
  [3]=>
  &string(4) "zero"
}
array(4) {
  [0]=>
  string(4) "zero"
  [1]=>
  string(3) "one"
  [2]=>
  string(3) "two"
  [3]=>
  &string(3) "one"
}
array(4) {
  [0]=>
  string(4) "zero"
  [1]=>
  string(3) "one"
  [2]=>
  string(3) "two"
  [3]=>
  &string(3) "two"
}
array(4) {
  [0]=>
  string(4) "zero"
  [1]=>
  string(3) "one"
  [2]=>
  string(3) "two"
  [3]=>
  &string(3) "two"
}

Answer 6

First loop

$v = $a[0];
$v = $a[1];
$v = $a[2];
$v = $a[3];

Yes! Current $v = $a[3] position.

Second loop

$a[3] = $v = $a[0], echo $v; // same as $a[3] and $a[0] == 'zero'
$a[3] = $v = $a[1], echo $v; // same as $a[3] and $a[1] == 'one'
$a[3] = $v = $a[2], echo $v; // same as $a[3] and $a[2] == 'two'
$a[3] = $v = $a[3], echo $v; // same as $a[3] and $a[3] == 'two'

because $a[3] is assigned by before processing.