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Date: Wed, 21 Oct 2009 22:47:41 -0400
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To: Richard K Miller <richardkmiller@gmail.com>
Subject: Re: [PHP-DEV] bug when using foreach with references?
From: marcot@tabini.ca (Marco Tabini)

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On 2009-10-21, at 10:40 PM, Richard K Miller wrote:

> I don't follow. Is this really the intended behavior? It seems quite =20=

> unintuitive that the original array would be modified by *empty* =20
> loops.

It is intended behaviour. Consider your code; at the end of this loop:

>>> $items =3D array('apple', 'banana', 'carrot');
>>> print_r($items);
>>> foreach ($items as &$item) { }
>>


$item will contain a reference to the last item of $items. Therefore, =20=

when you start the next loop:

>>> foreach ($items as $item) { }
>>


$item will receive each of the values stored in $items, in sequence. =20
But $item is a reference to the last item in $items, so whatever you =20
write into it will also be written into the last item of the array. =20
Thus, the loop will cause these values to be written there:

- On the first iteration, "apple"
- On the second iteration, "banana"
- On the third iteration, "banana" (because you are taking the third =20
item in the array, which you just overwrote)

>
> When would this behavior be desired?

Probably never, but that does not make it counterintuitive=97the =20
interpreter is behaving correctly based on the data it is handling and =20=

the instructions you are giving to it. The correct solution, as =20
someone has already recommended, is to unset the iterator if you want =20=

to reuse it.

Cheers,


Marco.=

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