Re: bug when using foreach with references?

Richard, enumerator exhausted, repeat your sub-sequence again you will get it e.g pointer and pointed Best Regards On Wed, Oct 21, 2009 at 7:01 PM, Richard K Miller <richardkmiller@gmail.com> wrote: > Hi Jill, > >>> <?php >>> $items = array('apple', 'banana', 'carrot'); >>> print_r($items); >>> foreach ($items as &$item) { } >> >> $item is now a reference to the last element of $items. >>> >>> print_r($items); >>> foreach ($items as $item) { } >> >> And here, the foreach loop writes to $item for each element: thus >> assiging that value to the last element of $items. > > The last element of $items is 'carrot'. Why does it print apple, banana, > banana? > > It sounds like you're saying that $item should be left with the value > 'carrot', which makes sense, but why would the original array be modified? > (These are empty foreach loops.) > >>> print_r($items); >>> ?> >>> >>> // Output: >>> Array >>> ( >>>    [0] => apple >>>    [1] => banana >>>    [2] => carrot >>> ) >>> Array >>> ( >>>    [0] => apple >>>    [1] => banana >>>    [2] => carrot >>> ) >>> Array >>> ( >>>    [0] => apple >>>    [1] => banana >>>    [2] => banana >>> ) >>> >>> Two bananas in the last set?! Not what I expected. >> >> You can fix your bug by using 'unset($item);' after the second >> foreach-loop. >> > > > > -- > PHP Internals - PHP Runtime Development Mailing List > To unsubscribe, visit: http://www.php.net/unsub.php > >