PHP copy array without references

Question

When I copy PHP array with reference, copy already has references from original

$arr = [1,2,3];
print_r($arr); echo"<br>";    
$x = &$arr[1];
$arr2 = $arr;
print_r($arr); print_r($arr2); echo"<br>";
$x = 8;
print_r($arr); print_r($arr2); echo"<br>";

Result:

Array ( [0] => 1 [1] => 2 [2] => 3 ) 
Array ( [0] => 1 [1] => 2 [2] => 3 ) Array ( [0] => 1 [1] => 2 [2] => 3 ) 
Array ( [0] => 1 [1] => 8 [2] => 3 ) Array ( [0] => 1 [1] => 8 [2] => 3 )

How can I copy an array, so it has not changed with the original reference?

Array ( [0] => 1 [1] => 2 [2] => 3 ) 
Array ( [0] => 1 [1] => 2 [2] => 3 ) Array ( [0] => 1 [1] => 2 [2] => 3 ) 
Array ( [0] => 1 [1] => 8 [2] => 3 ) Array ( [0] => 1 [1] => 2 [2] => 3 )

Possible duplicate of php copying array elements by value, not by reference — Muhammad Saqlain
– Muhammad Saqlain, Commented Feb 19, 2017 at 19:27

mickmackusa · Accepted Answer · 2017-03-11 23:51:04Z

0

If your issue is solved using the duplicate link in your question's first comment (supported by 4 others at the time I'm writing this). Please delete your question so that SO can reduce duplicate questions / needless bloat.

Otherwise, just declare a static copy of the original array for future use.

$arr = [1,2,3];
$copy=$arr;  // preserve the original
print_r($arr); echo"<br>";    
$x = &$arr[1];
$arr2 = $arr;
print_r($arr); print_r($arr2); echo"<br>";
$x = 8;
print_r($arr); print_r($copy); echo"<br>";

answered Mar 11, 2017 at 23:51

mickmackusa♦

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