Re[2]: [PHP] Bug ?

David, thank you very much. I suspected smth. like this, but still, it is weird: PHP already has the "&" operator (?) for assigning by reference. And there is more: I noticed that if I use the "sort" function instead of "array_multisort", it works as expected. I still think there is a bug involved :( >>I tried something like this: >> >><? >>$arrayA = array(5,4,3,2,1); >>$arrayB = $arrayA; >>array_multisort($arrayA); >>echo $arrayA[0], "<br>"; >>echo $arrayB[0]; >>?> >> >> >>The output is: >>1 >>1 >> >>I think it should be: >>1 >>5 DO> It's not a bug (ie this is expected behaviour in PHP4 for various DO> sensible reasons), but it can sometimes throw up odd effects if you're DO> not expecting it. DO> What you're running into here is the difference between a deep copy DO> (make a copy of a piece of memory) and a shallow copy (make two DO> variables point to the same piece of memory). DO> The way to think of it is that $arrayA doesn't actually contain your DO> array values - it contains a reference (a pointer in C-speak) to the DO> memory location where the array values are stored. The line DO> $arrayB = $arrayA; DO> is copying the reference, not the values. You can run up against this DO> behaviour in quite a few of the post-C++ languages, and it can be DO> disconcerting if you're used to languages where copies are all deep DO> unless flagged otherwise. DO> To add insult to injury, some array operations can implicitly cause a DO> deep copy to be made. Try this, which adds one extra line: DO> <? DO> function echo_array($a) { DO> foreach ($a as $v) { DO> echo ($v); DO> echo (', '); DO> } DO> echo ("<br>"); DO> } DO> $arrayA = array(5,4,3,2,1); DO> $arrayB = $arrayA; DO> $arrayA[] = 6; // this line added DO> array_multisort($arrayA); DO> echo_array($arrayA); DO> echo_array($arrayB); ?>> DO> More details (maybe) here: DO> http://www.zend.com/zend/art/ref-count.php -- Ciprian > Un cuvant de sfarsit: > A cynic knows the price of everything & value of nothing