Jochem Maas wrote: >>> foo($a = 5); > > by definition the expression is evaluated _before_ the function is > called - so the expression is not passed to the function, the result > of the expression is passed ... I was under the impression that the > the expression evaluates to a 'pointer' (I'm sure thats bad > terminology) to $a ... which can taken by reference by the function. > > possibly I am completely misunderstanding what goes on here.
When used as an expression, an assignment evaluates to whatever is on the right side of the assignment operator, not the left. Example: var_dump($a = 5); outputs int(5) var_dump($a = "some string"); outputs string(11) "some string" So, as far as foo() knows: foo($a = 5); and foo(5); are exactly the same... -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
