> I think, and I could be completely wrong, that copying a variable actually > creates a reference. The data is only copied when the variable referenced is > modified. > That's true. What I left out of my explanation (in order to keep it simple) is that when you "copy" a variable, a new label is created to point to the same zval, and the zval's refcount is incrmented but the is_ref flag is *not* set (I referred to this offhand as non-reference manner of multiple labels referring to the same value). Then when one of the referring labels says "I want to change my `copy` of the data." It notices that someone else is also referring to this value (in a non-reference manner) and "separates" the zval: This amounts to making a true copy of the zval (with a refcount of 1, and an is_ref of 0) and decrements the refcount of the original zval (since one fewer label is referring to it). This is the process known as "copy on change".
$foo = 1; /* $foo (label) ------> 1(value) (is_ref:0 refcount:1) */ $bar = $foo; /* $foo (label) -------> 1(value) */ /* $bar (label) ---/ (is_ref:0 refcount:2) */ $foo = 2; /* $bar (label) -------> 1(value) (is_ref:0 refcount:1) */ /* $foo (label) -------> 2(value) (is_ref:0 refcount:1) */ -Sara Ya just had to make it complicated didn't you. -- PHP Internals - PHP Runtime Development Mailing List To unsubscribe, visit: http://www.php.net/unsub.php
