note 76272 deleted from language.oop5.basic by felipe

Note Submitter: Stephen ---- 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.