Doc #62383 [Opn->Csd]: array_replace() arguments not passed by reference

Edit report at https://bugs.php.net/bug.php?id=62383&edit=1 ID: 62383 Updated by: salathe@php.net Reported by: confidential99 at gmail dot com Summary: array_replace() arguments not passed by reference -Status: Open +Status: Closed Type: Documentation Problem Package: Arrays related PHP Version: Irrelevant -Assigned To: +Assigned To: salathe Block user comment: N Private report: N New Comment: This bug has been fixed in SVN. Snapshots of the sources are packaged every three hours; this change will be in the next snapshot. You can grab the snapshot at http://snaps.php.net/. For Windows: http://windows.php.net/snapshots/ Thank you for the report, and for helping us make PHP better. The arguments for array_replace() and array_replace_recursive() were incorrectly marked as being passed by reference. Thanks for spotting this. Previous Comments: ------------------------------------------------------------------------ [2012-06-22 21:42:04] salathe@php.net Automatic comment from SVN on behalf of salathe Revision: http://svn.php.net/viewvc/?view=revision&amp;revision=326300 Log: array_replace(_recursive) parameters are not by reference (doc #62383) ------------------------------------------------------------------------ [2012-06-22 19:36:09] crmalibu at gmail dot com Ok, so capturing the return value of the function is the proper way to use this function. However, I agree with the original bug submitter that it seems very odd that the function is documented as accepting arguments by reference, yet it's functionality doesn't *need* args by ref because it will *never* modify any of the args. I think any reasonable person who reads the php documentation would assume that an argument documented as a reference, has some reason for being a reference. And that reason would be that it may be modified. But this function won't ever modify an arg, and so it shouldn't be documented as such. ------------------------------------------------------------------------ [2012-06-22 10:53:08] fd@php.net You have to use $array1 = array_replace($array1,$array2,$array3); if you want your expected result. This is not a bug but the expected behaviour as stated in the documentation for array_replace(). ------------------------------------------------------------------------ [2012-06-21 14:27:17] confidential99 at gmail dot com Description: ------------ The document for the array_replace function says that "array_replace() replaces the values of the first array with the same values from all the following arrays...." The signature for the function indicates the arrays will be passed by reference. Test script: --------------- <?php $array1 = array('AMZN'=>223.02,'BA'=>8.14,'SAN'=>6.26,'AAPL'=>586.05); $array2 = array('AMZN'=>223.84,'BA'=>8.11,'SAN'=>6.40,'FB'=>31.60,'ORCL'=>28.49); $array3 = array('AMZN'=>224.25,'BA'=>8.10,'ORCL'=>28.39,'INTC'=>27.38); array_replace($array1,$array2,$array3); var_dump($array1); Expected result: ---------------- array(7) { ["AMZN"]=> float(224.25) ["BA"]=> float(8.1) ["SAN"]=> float(6.4) ["AAPL"]=> float(586.05) ["FB"]=> float(31.6) ["ORCL"]=> float(28.39) ["INTC"]=> float(27.38) } Actual result: -------------- array(4) { ["AMZN"]=> float(223.02) ["BA"]=> float(8.14) ["SAN"]=> float(6.26) ["AAPL"]=> float(586.05) } ------------------------------------------------------------------------ -- Edit this bug report at https://bugs.php.net/bug.php?id=62383&edit=1