PHP unreference

Question

Lets say i have this code:

$val = 1;
$arr = Array();
$arr['val'] =& $val;
$val = 2;
echo $arr['val'];

This will print out 2, because $val was passed to $arr by reference.

My question is : if i passed a value to an array by reference, is there a way to remove that reference later on, making it a simple copied value ? To make it clearer, i would like something like this:

$val = 1;
$arr = Array();
$arr['val'] =& $val;

$arr['val'] = clone $arr['val']; 
// Or better yet:
$arr = clone $arr;

$val = 2;
echo $arr['val'];

And this should print out 1 (because the array was cloned, before the referenced variable changed). Howerver, clone does not work with arrays, it only works with objects.

Any ideas? I really have no clue how to do this. I tried writing a recursive copy function, but that didn't work.

user229044 · Accepted Answer · 2011-02-09 13:05:59Z

4

You could unset the index and then reassign by-value instead of by reference.

unset($arr['val']);
$arr['val'] = $val;

answered Feb 9, 2011 at 13:05

user229044♦

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2 Comments

Kipras Over a year ago

And how about, if i don't have the original $val ? Say i'm now in a different function and i only have access to $arr, and i want to remove all inner references.

user229044 Over a year ago

@Kipras Then store the value, unset the array index, and reassign

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