maybe you should be foreach()ing with references?
php.net : search "foreach" :
As of PHP 5, you can easily modify array's elements by preceding
$value with &. This will assign reference instead of copying the
value.
<?php
$arr = array(1, 2, 3, 4);
foreach ($arr as &$value) {
$value = $value * 2;
}
// $arr is now array(2, 4, 6, 8)
unset($value); // break the reference with the last element
?>
This is possible only if iterated array can be referenced (i.e. is variable),
On Tue, Mar 16, 2010 at 5:43 PM, Robert Cummings <robert@xxxxxxxxxxxxx> wrote:
> Peter Lind wrote:
>>
>> Hmm, will probably have to look inside PHP for this ... the foreach
>> loop will copy each element as it loops over it (without actually
>> copying, obviously), however there's no change happening to the
>> element at any point and so there's nothing to suggest to the
>> copy-on-write to create a new instance of the sub-array.
>>
>> It should look like this:
>> $a = array(0, 1, 2, array(0, 1, 2, 3), 4, 5, 6, .... n);
>> $b = $a[3];
>> doStuffs($b);
>>
>> Whether or not you loop over $a and thus move the internal pointer,
>> you don't change (well, shouldn't, anyway) $b as that's a subarray
>> which has it's own internal pointer, that isn't touched.
>>
>> Or maybe I've gotten this completely backwards ...
>
> I'm not sure of the exact reason... PHP has the following comment:
>
> Note: Unless the array is referenced, foreach operates on a copy
> of the specified array and not the array itself. foreach
> has some side effects on the array pointer. Don't rely on
> the array pointer during or after the foreach without
> resetting it.
>
> http://php.net/manual/en/control-structures.foreach.php
>
> I've always found the foreach loop to a be a bit on the bizarre size, this
> is probably just another one of those bizarrities :)
>
> Cheers,
> Rob.
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