Re: Re: unsetting a referenced parameter in a function

On Wed, Jul 22, 2009 at 9:27 PM, Tom Worster <fsb@thefsb.org> wrote: > On 7/22/09 6:09 PM, "Shawn McKenzie" <nospam@mckenzies.net> wrote: > > > Tom Worster wrote: > >> though the manual is perfectly clear that this should be expected, i was > a > >> bit surprised that the result of the following is 42 > >> > >> <?php > >> function foo(&$a) { > >> $a = 42; > >> unset($a); > >> $a = 'meaning'; > >> } > >> foo($a); > >> print("$a\n"); > >> ?> > >> > >> normally i would expect unset() to free some memory. but in this example > it > >> doesn't and has a different behavior: it releases foo's reference to the > >> global $a, allowing the next line to define a local $a. > >> > >> i think i'd have preferred compile error. > >> > >> > > > > Well, you unset the reference and then you assigned 'meaning' to a local > > function variable $a. Why would you get a compile error? > > when you state it in those terms (which are clearly correct) i wouldn't. > > but if the way i think is "unset() destroys the specified variables" (as > the > manual puts it) then i expect that the specified variable would be > destroyed, not the reference. > > so, as i said, i was a bit surprised when the variable wasn't destroyed. > once i understood what was happening, i thought it a bit confusing to have > such scope-dependent differences in behavior of a language element. > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > RTM: http://php.net/unset The behavior of *unset()* inside of a function can vary depending on what type of variable you are attempting to destroy. If a globalized variable is *unset()* inside of a function, only the local variable is destroyed. The variable in the calling environment will retain the same value as before *unset()* was called. / ... / If a variable that is PASSED BY REFERENCE is *unset()* inside of a function, only the local variable is destroyed. The variable in the calling environment will retain the same value as before *unset()* was called. I think the manual is very clear about unset. -- Martin Scotta