Is there a function to make a copy of a PHP array to another?

Question

Is there a function to make a copy of a PHP array to another?

I have been burned a few times trying to copy PHP arrays. I want to copy an array defined inside an obj

Answer 1

Safest and cheapest way I found is:

<?php 
$b = array_values($a);

This has also the benefit to reindex the array.

This will not work as expected on associative array (hash), but neither most of previous answer.

Answer 2

Since this wasn't covered in any of the answers and is now available in PHP 5.3 (assumed Original Post was using 5.2).

In order to maintain an array structure and change its values I prefer to use array_replace or array_replace_recursive depending on my use case.

http://php.net/manual/en/function.array-replace.php

Here is an example using array_replace and array_replace_recursive demonstrating it being able to maintain the indexed order and capable of removing a reference.

http://ideone.com/SzlBUZ

The code below is written using the short array syntax available since PHP 5.4 which replaces array() with []. http://php.net/manual/en/language.types.array.php

Works on either offset indexed and name indexed arrays

$o1 = new stdClass;
$a = 'd';
//This is the base array or the initial structure
$o1->ar1 = ['a', 'b', ['ca', 'cb']];
$o1->ar1[3] = & $a; //set 3rd offset to reference $a

//direct copy (not passed by reference)
$o1->ar2 = $o1->ar1; //alternatively array_replace($o1->ar1, []);
$o1->ar1[0] = 'z'; //set offset 0 of ar1 = z do not change ar2
$o1->ar1[3] = 'e'; //$a = e (changes value of 3rd offset to e in ar1 and ar2)

//copy and remove reference to 3rd offset of ar1 and change 2nd offset to a new array
$o1->ar3 = array_replace($o1->ar1, [2 => ['aa'], 3 => 'd']);

//maintain original array of the 2nd offset in ar1 and change the value at offset 0
//also remove reference of the 2nd offset
//note: offset 3 and 2 are transposed
$o1->ar4 = array_replace_recursive($o1->ar1, [3 => 'f', 2 => ['bb']]);

var_dump($o1);

Output:

["ar1"]=>
  array(4) {
    [0]=>
    string(1) "z"
    [1]=>
    string(1) "b"
    [2]=>
    array(2) {
      [0]=>
      string(2) "ca"
      [1]=>
      string(2) "cb"
    }
    [3]=>
    &string(1) "e"
  }
  ["ar2"]=>
  array(4) {
    [0]=>
    string(1) "a"
    [1]=>
    string(1) "b"
    [2]=>
    array(2) {
      [0]=>
      string(2) "ca"
      [1]=>
      string(2) "cb"
    }
    [3]=>
    &string(1) "e"
  }
  ["ar3"]=>
  array(4) {
    [0]=>
    string(1) "z"
    [1]=>
    string(1) "b"
    [2]=>
    array(1) {
      [0]=>
      string(2) "aa"
    }
    [3]=>
    string(1) "d"
  }
  ["ar4"]=>
  array(4) {
    [0]=>
    string(1) "z"
    [1]=>
    string(1) "b"
    [2]=>
    array(2) {
      [0]=>
      string(2) "bb"
      [1]=>
      string(2) "cb"
    }
    [3]=>
    string(1) "f"
  }

Answer 3

In php array, you need to just assign them to other variable to get copy of that array. But first you need to make sure about it's type, whether it is array or arrayObject or stdObject.

For Simple php array :

$a = array(
'data' => 10
);

$b = $a;

var_dump($b);

output:

array:1 [
  "data" => 10
]

Answer 4

<?php
function arrayCopy( array $array ) {
        $result = array();
        foreach( $array as $key => $val ) {
            if( is_array( $val ) ) {
                $result[$key] = arrayCopy( $val );
            } elseif ( is_object( $val ) ) {
                $result[$key] = clone $val;
            } else {
                $result[$key] = $val;
            }
        }
        return $result;
}
?>

Answer 5

In PHP arrays are assigned by copy, while objects are assigned by reference. This means that:

$a = array();
$b = $a;
$b['foo'] = 42;
var_dump($a);

Will yield:

array(0) {
}

Whereas:

$a = new StdClass();
$b = $a;
$b->foo = 42;
var_dump($a);

Yields:

object(stdClass)#1 (1) {
  ["foo"]=>
  int(42)
}

You could get confused by intricacies such as ArrayObject, which is an object that acts exactly like an array. Being an object however, it has reference semantics.

Edit: @AndrewLarsson raises a point in the comments below. PHP has a special feature called "references". They are somewhat similar to pointers in languages like C/C++, but not quite the same. If your array contains references, then while the array itself is passed by copy, the references will still resolve to the original target. That's of course usually the desired behaviour, but I thought it was worth mentioning.

Answer 6

PHP will copy the array by default. References in PHP have to be explicit.

$a = array(1,2);
$b = $a; // $b will be a different array
$c = &$a; // $c will be a reference to $a