#41474 [Opn->WFx]: Clarification of array assignment behaviour
| From: | vrana@php.net | Date: | Fri, 17 Aug 2007 08:12:09 +0000 |
| Subject: | #41474 [Opn->WFx]: Clarification of array assignment behaviour | ||
| References: | 1 | Groups: | php.doc |
| Request: | Send a blank email to phpdoc+get-969377532@lists.php.net to get a copy of this message | ||
ID: 41474
Updated by: vrana@php.net
Reported By: tayloj1 at uk dot ibm dot com
-Status: Open
+Status: Wont fix
Bug Type: Documentation problem
PHP Version: Irrelevant
New Comment:
Is has nothing to do with arrays - that's the way how assignment works
even with normal variables.
Previous Comments:
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[2007-05-23 12:09:14] tayloj1 at uk dot ibm dot com
Description:
------------
Ref: http://uk.php.net/manual/en/language.types.array.php
Current PHP manual says:
"You should be aware that array assignment always involves value
copying. It also means that the internal array pointer used by current()
and similar functions is reset. You need to use the reference operator
to copy an array by reference. "
This statement can be misunderstood and should be clarified as follows:
" You should be aware that array assignment <delete word 'always'>
involves value copying. It also means that the internal array pointer
used by current() and similar functions is reset. You need to use the
reference operator to copy an array by reference. When the array is
copied by value, elements in the array are copied by reference if they
are already references. An element in an array will be a reference if it
was created by a reference assignment or if a reference has been created
to the element."
Example:
<?php
$a = array (1,2) ;
$b = &$a[0] ; // make $a[0] part of ref set
$c = $a ;
$a[0] = 3 ;
$a[1] = 4 ;
echo $c[0] ; // outputs 3 because $c[0] was copied by reference
echo "\n" ;
echo $c[1] ; // outputs 2 (not 4) because $c[1] was copied by value
?>
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Edit this bug report at http://bugs.php.net/?id=41474&edit=1