On Thu, Mar 15, 2012 at 5:22 PM, Patrick ALLAERT <patrickalla...@php.net> wrote: > 2012/3/15 Nikita Popov <nikita....@googlemail.com>: >> If I am understanding the text correctly it is saying that >> $f1 = f1(); >> $f2 = f2($f1); >> $f3 = f3($f2); >> is using more memory than >> $f3 = f3(f2(f1())); >> >> For me this doesn't make any sense. In the latter case PHP will also >> create temporary variables to store the return values. There should be >> no difference in memory consumption. > > It does make sense to me. > > In the first case, when calling f3(), $f1 is still referenced. > In the second case, when calling f3(), the result of f2() is > referenced, but there is no more active reference to the result of > f1(). I don't really know when PHP frees temporary variables, but my guess was that they are freed when the scope is left.
If that is not true, then forget whatever I said. But if it is true, then there is no inherent difference between the two version. The only difference is that explicit $variables would need an entry in the active symbol table, which is pretty much negligible. -- PHP Internals - PHP Runtime Development Mailing List To unsubscribe, visit: http://www.php.net/unsub.php
