note 78799 added to language.references.pass

If you intend to pass a copy of an object to a function, then you should use 'clone' to create a copy explicity. In PHP5, objects appear to always be passed by reference (unlike PHP4), but this is not strictly true. The way I think of it is that if you use '=&' (or you explicitly pass to a function by reference) the variable behaves like a C++ reference, except that you can re-assign the reference to something else (which is not possible in C++). When you use '=' with an object, it is more like you are dealing with a pointer (if you think in this way, the '->' access element through pointer operator has the same behaviour in C++). <?php class z { public $var = ''; } function f(&$obj1, $obj2, $obj3, $obj4) { $obj1->var = null; $obj2->var = null; $obj3 = new z(); $obj3->var = null; $obj4 = clone $obj4; $obj4->var = null; } $a = new z(); $b = new z(); $c = new z(); $d = new z(); f($a, $b, $c, $d); var_dump($a); // object(z)#1 (1) { ["var"] => NULL } var_dump($b); // object(z)#2 (1) { ["var"] => NULL } var_dump($c); // object(z)#3 (1) { ["var"] => string(0) "" } var_dump($d); // object(z)#4 (1) { ["var"] => string(0) "" } ?> Stephen 08-Jul-2007 04:54 jcastromail at yahoo dot es stated: **** in php 5.2.0 for classes $obj1 = $obj2; is equal to $obj1 = &$obj2;" **** However, that is not completely true. While both = and =& will make a variable refer to the same object as the variable being assigned to it, the explicit reference assignment (=&) will keep the two variables joined to each other, whereas the assignment reference (=) will make the assigned variable an independent pointer to the object. An example should make this clearer: <?php class z { public $var = ''; } $a = new z(); $b =& $a; $c = $a; $a->var = null; var_dump($a); print '<br>'; var_dump($b); print '<br>'; var_dump($c); print '<br><br>'; $a = 2; var_dump($a); print '<br>'; var_dump($b); print '<br>'; var_dump($c); print '<br><br>'; ?> This outputs: object(z)#1 (1) { ["var"]=> NULL } object(z)#1 (1) { ["var"]=> NULL } object(z)#1 (1) { ["var"]=> NULL } int(2) int(2) object(z)#1 (1) { ["var"]=> NULL } So although all 3 variables reflect changes in the object, if you reassign one of the variables that were previously joined by reference to a different value, BOTH of those variables will adopt the new value. Perhaps this is because =& statements join the 2 variable names in the symbol table, whereas = statements applied to objects simply create a new independent entry in the symbol table that simply points to the same location as other entries. I don't know for sure - I don't think this behavior is documented in the PHP manual, so perhaps somebody with more knowledge of PHP's internals can clarify what is going on. ---- Server IP: 64.71.164.2 Probable Submitter: 81.5.171.23 ---- Manual Page -- http://www.php.net/manual/en/language.references.pass.php Edit -- https://master.php.net/note/edit/78799 Del: integrated -- https://master.php.net/note/delete/78799/integrated Del: useless -- https://master.php.net/note/delete/78799/useless Del: bad code -- https://master.php.net/note/delete/78799/bad+code Del: spam -- https://master.php.net/note/delete/78799/spam Del: non-english -- https://master.php.net/note/delete/78799/non-english Del: in docs -- https://master.php.net/note/delete/78799/in+docs Del: other reasons-- https://master.php.net/note/delete/78799 Reject -- https://master.php.net/note/reject/78799 Search -- https://master.php.net/manage/user-notes.php