Duplicate array but maintain pointer links

Question

Suppose I have an array of nodes (objects). I need to create a duplicate of this array that I can modify without affecting the source array. But changing the nodes will affect the source nodes. Basically maintaining pointers to the objects instead of duplicating their values.

// node(x, y)
$array[0] = new node(15, 10);
$array[1] = new node(30, -10);
$array[2] = new node(-2, 49);

// Some sort of copy system
$array2 = $array;

// Just to show modification to the array doesn't affect the source array
array_pop($array2);
if (count($array) == count($array2))
  echo "Fail";    

// Changing the node value should affect the source array
$array2[0]->x = 30;
if ($array2[0]->x == $array[0]->x)
   echo "Goal";

What would be the best way to do this?

I don't get it. How can you modify the array without affecting the source array when you want it to update the source array when you alter a value in the duplicated array? Can you give an example of what you are doing with this? — Bryan Denny, Mar 24 '10 at 15:07
I think he means that the duplicated array should have the references to the same instances as the first array, but be itself indepent from array. So changing the contents of the second array won't modify the first array, but changing some parameter in a class referenced by the second array will modify the same instance that is refenced by the first array. — Kjir, Mar 24 '10 at 15:12
The code already has the behavior you desire. Well, unless you're using php 4. — goat, Mar 24 '10 at 15:26
In my test your code works as you say it should... (php 5.3) It seems the array is just a copy, but its elements are assigned by reference automatically — Nicolò Martini, Mar 24 '10 at 15:28

score 2 · Accepted Answer · edited May 23 '17 at 11:48

2

If you use PHP 5:

Have you run your code? It is already working, no need to change anything. I get:

Goal

when I run it.

Most likely this is because the values of $array are already references.

Read also this question. Although he OP wanted to achieve the opposite, it could be helpful to understand how array copying works in PHP.

Update:

This behaviour, when copying arrays with objects, the reference to the object is copied instead the object itself, was reported as a bug. But no new information on this yet.

If you use PHP 4:

(Why do you still use it?)

You have to do something like:

$array2 = array();

for($i = 0; $i<count($array); $i++) {
    $array2[$i] = &$array[$i];
}

edited May 23 '17 at 11:48

Community

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answered Mar 24 '10 at 15:12

Felix Kling

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Ahhh, I didn't actually test the code. I was expecting having to use `copy` or something like that. Thanks for the info. How does this work when passing through functions? – St. John Johnson Mar 24 '10 at 15:34
@StJohn Johnson: What do you mean with passing through functions? An object passed to a function as parameter? This should also be passed by reference (in PHP 5). See also here: http://devzone.zend.com/article/1714 – Felix Kling Mar 24 '10 at 15:41

score 0 · Answer 2 · answered Mar 24 '10 at 15:16

0

it is some time that I don' write PHP code, but does the code

// Some sort of copy system
$array2 = $array;

actually work?

Don't you have to copy each element of the array in a new one?

answered Mar 24 '10 at 15:16

Giovanni Di Milia

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No, you don't have to copy every element. The documentation says: *Array assignment always involves value copying.* http://php.net/manual/en/language.types.array.php – Felix Kling Mar 24 '10 at 15:18
you are right... during my answer I was thinking how to keep pointers instead of a copy of the objects... – Giovanni Di Milia Mar 24 '10 at 15:23

Duplicate array but maintain pointer links

2 Answers2