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[2011-12-06 10:05 UTC] developer at elementica dot com
Description: ------------ On a standard installation of PHP 5.3.x on many OSs (including Linux and Windows) we: - set an array with at least two elements - loop it using foreach by-reference changing each element - create a copy of the array in another one using the "=" assignment - loop on the copy as before - we find the original array changed (usually or always: last element) Example code given. Documentation says "Array assignment always involves value copying. Use the reference operator to copy an array by reference." (http://php.net/manual/en/language.types.array.php). It seems a problem of "foreach" used with the "by-reference" syntax. Could be related to https://bugs.php.net/bug.php?id=8130 (subject appears to be the same, but here we have a foreach problem, while nothing is stated in #8130 about it). Test script: --------------- $array1 = array(0 => 'zero', 1 => 'one'); foreach ($array1 as &$elem) { $elem .= ' (1)'; }; // NOTE: if we use $array1[0] .= ' (1)'; $array1[1] .= ' (1)'; no problem arises $array2 = $array1; foreach ($array2 as &$elem) { $elem .= ' (2)'; }; // NOTE: if we use $array2[0] .= ' (1)'; $array2[1] .= ' (1)'; the problem is still here, so first foreach seems to be the point print_r ($array1); // BUG! EXPECTED: array(0 => 'zero (1)', 1 => 'one (1)'), OUTPUT: array(0 => 'zero (1)', 1 => 'one (1) (2)') Expected result: ---------------- Array ( [0] => zero (1) [1] => one (1) ) Actual result: -------------- Array ( [0] => zero (1) [1] => one (1) (2) ) PatchesPull RequestsHistoryAllCommentsChangesGit/SVN commits
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Last updated: Wed Apr 23 18:01:28 2025 UTC |
I knew the trick. But I think that's just a trick (thankx very much, anyway!) Foreach takes two arguments, the second being the iteration variable that should have a scope relative to the loop. This could also be seen as a feature, but we also find stated that "Referencing $value is only possible if the iterated array can be referenced (i.e. if it is a variable)". So: - if it's to be seen as a feature the latter shouldn't be true - if it isn't (as it seems better) unset shouldn't be necessary Let's have a different example... Considering a function we could have: function fun(&$x) { $x++; }; $var1 = 7; // this is like the $array1 assignment fun($var1); // this is like the foreach loop for $array1 (*) $var2 = $var1; // this is like the $array2=$array1 assignment fun($var2); // this is like the foreach loop for $array2 (*) print_r($var1); // WE FIND 8 If we look at "fun" as "foreach" it seems to me we're in a similar case, but here $var1 isn't affected by the change of $var2. (*) just like before... we're applying a construct (foreach before, a function now) to a referenced variable. $x is just a formal parameter, just like $elem I think the point is here... is $elem a formal parameter or not? I think yes would be a better answer.