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
– Bryan Denny, Commented Mar 24, 2010 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
– Kjir, Commented Mar 24, 2010 at 15:12

Community · Accepted Answer · 2017-05-23 11:48:41Z

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, 2017 at 11:48

CommunityBot

11 silver badge

answered Mar 24, 2010 at 15:12

Felix Kling

820k181 gold badges1.1k silver badges1.2k bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

St. John Johnson Over a year ago

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?

Felix Kling Over a year ago

@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: devzone.zend.com/article/1714

Giovanni Di Milia · Accepted Answer · 2010-03-24 15:16:21Z

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, 2010 at 15:16

Giovanni Di Milia

14.1k13 gold badges58 silver badges71 bronze badges

2 Comments

Felix Kling Over a year ago

No, you don't have to copy every element. The documentation says: Array assignment always involves value copying. php.net/manual/en/language.types.array.php

Giovanni Di Milia Over a year ago

you are right... during my answer I was thinking how to keep pointers instead of a copy of the objects...

Collectives™ on Stack Overflow

Duplicate array but maintain pointer links

2 Answers 2

2 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related