PHP: making a copy of a reference variable

Question

If I make a copy of a reference variable. Is the new variable a pointer or does it hold the value of the variable the pointer was referring to?

References are not pointers. See http://php.net/manual/en/language.references.arent.php — Artefacto, May 15 '10 at 13:06
Sometimes you must use them, for instance when implementing offsetget. Though no doubt that references in PHP are a mess. — Artefacto, May 15 '10 at 13:49

score 8 · Answer 1 · answered May 15 '10 at 12:42

8

Let's make a quick test:

<?php

$base = 'hello';
$ref =& $base;
$copy = $ref;

$copy = 'world';

echo $base;

Output is hello, therefore $copy isn't a reference to %base.

answered May 15 '10 at 12:42

Crozin

2

Which should be expected - the point of `$ref` being a reference to `$base` is that `$copy = $ref` should have the exact same effect as `$copy = $base`. – Tgr May 15 '10 at 13:06
This made me understand references far better. For some reason i though that to copy a reference you could `$copy = $ref;`. Now i understand that is like a dereference and copies the data that the reference references. I think using C pointers makes this confusing as you have to dereference manually. – Lightbulb1 Jul 02 '14 at 13:46
@Lightbulb1: That's because PHP's references are more like C++ references than C/C++ pointers. ;) – Crozin Jul 02 '14 at 14:22
Pointers and references are very confusing. I think i'm getting there haha – Lightbulb1 Jul 02 '14 at 15:14

score 8 · Accepted Answer · edited Sep 28 '21 at 08:47

8

It holds the value. If you wanted to point, use the & operator to copy another reference:

$a = 'test';
$b = &$a;
$c = &$b;

edited Sep 28 '21 at 08:47

AlexSp3

answered May 15 '10 at 12:47

Delan Azabani

1

Don't say "point". he'll think there's some difference between $a, $b and $c (like a pointer and a pointed). – Artefacto May 15 '10 at 13:51

score -1 · Answer 3 · answered Aug 21 '15 at 10:09

-1

Let me murk the water with this example:

$a = array (1,2,3,4);
foreach ($a as &$v) {

}
print_r($a);

foreach ($a as $v) {
  echo $v.PHP_EOL;
}
print_r($a);

Output:

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

1
2
3
3

Array
(
   [0] => 1
   [1] => 2
   [2] => 3
   [3] => 3
)

answered Aug 21 '15 at 10:09

SubjectX

1

Your example is **intentionally confusing**. after `foreach ($anything as &$v) {}` variable `$v` will hold the reference to last array element (while this is confusing, it's correct behavior), you should either `unset($v)` or keep in mind that `$v` is a reference. **Pro-tip: use var_dump() instead of print_r(), it will show &references in arrays** – Sanya_Zol Feb 18 '17 at 20:36

3 Answers3