ID: 39811
User updated by: wishm at bk dot ru
Reported By: wishm at bk dot ru
Status: Bogus
Bug Type: Arrays related
Operating System: win32
PHP Version: 4.4.4
New Comment:
I think you dont understand the purpose of the array_diff() function.
The element is considered present if and only if its' VALUE is equal to
the same-named element's VALUE in the second array! Try
array_diff(array("a"=>"1"),array("a"=>"2")). The element a = 1 does not
present at second array so it will be returned. You didnt even tried to
understand the essence of the bug! It is 100% PHP error when 2 is
considered equal to ANY value under conditions I've described. Please
fix that!
Previous Comments:
------------------------------------------------------------------------
[2006-12-13 10:06:39] [EMAIL PROTECTED]
http://php.net/array_diff
array_diff() returns an array containing **all the values of array1
that are not present in any of the other arguments*.
In your case all values of $a are present in $b.
------------------------------------------------------------------------
[2006-12-13 07:49:54] wishm at bk dot ru
Description:
------------
array_diff() function fails on comparing any array element with 2, when
another pair of elements are both set to 2. Fails with that settings
only.
Reproduce code:
---------------
<?php
$a = array("2","2");
$b = array("2","any_symbols_here");
// same thing on $b = array("any_symbols_here","2");
print_r(array_diff($a,$b));
?>
Expected result:
----------------
Array ( [1] => 2 )
// second elements are not equal so function should return that element
from $a array
Actual result:
--------------
Array ( )
//empty array - function failed
------------------------------------------------------------------------
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Edit this bug report at http://bugs.php.net/?id=39811&edit=1
