What about objects?
class Foo {
public $foo;
}
function test($o) {
$o->foo->foo->foo = 2;
}
$bar = new Foo;
$bar->foo = new Foo;
$bar->foo->foo = new Foo;
test( $bar );
---
Also... is it better to pass an object as a parameter rather than many
values?
function withValues($anInteger, $aBool, $aString) {
var_dump($anInteger, $aBool, $aString);
}
function withObject(ParamOject $o) {
var_dump( $o->theInteger(), $o->theBool(), $o->theString() );
}
Martin Scotta
On Wed, Jan 19, 2011 at 5:03 AM, Hannes Landeholm <landeh...@gmail.com>wrote:
> Using references does not speed up PHP. It does that already
> internally, if I'm not mistaken. The point of my post was that
> assigning values to tree arrays are in general faster than a full
> array copy.
>
> Hannes
>
> On 19 January 2011 08:36, Ben Schmidt <mail_ben_schm...@yahoo.com.au>
> wrote:
> > Yep. PHP does clock up memory very quickly for big arrays, objects with
> lots
> > of members and/or lots of small objects with large overheads. There are a
> > LOT of zvals and zobjects and things around the place, and their overhead
> > isn't all that small.
> >
> > Of course, if you go to the trouble to construct arrays using references,
> > you can avoid some of that, because a copy-on-write will just copy the
> > reference. It does mean you're passing references, though.
> >
> > $bar['baz'] = 1;
> > $poink['narf'] = 1;
> > $a['foo']['bar'] =& $bar;
> > $a['foo']['poink'] =& $poink;
> >
> > Then if you test($a), $bar and $poink will be changed, since they are
> > 'passed by reference'--no copying needs to be done. It's almost as if $b
> > were passed by reference, but setting $b['blip'] wouldn't show up in $a,
> > because $a itself would be copied in that case, including the references,
> > which would continue to refer to $bar and $poink. So a much quicker copy,
> > but obviously not the same level of isolation that you might expect or
> > desire. Unless you did some jiggerypokery like $b_bar=$b['bar'];
> > $b['bar']=$b_bar; which would break the reference and make a copy of just
> > that part of the array. But this is a pretty nasty caller-callee
> > co-operative kind of thing. Just a thought to throw into the mix, though.
> >
> > Disclaimer: I'm somewhat out of my depth here. But I'm sure someone will
> > jump on me if I'm wrong.
> >
> > Ben.
> >
> >
> >
> > On 19/01/11 6:09 PM, Larry Garfield wrote:
> >>
> >> That's what I was afraid of. So it does copy the entire array. Crap.
> :-)
> >>
> >> Am I correct that each level in the array represents its own ZVal, with
> >> the
> >> additional memory overhead a ZVal has (however many bytes that is)?
> >>
> >> That is, the array below would have $a, foo, bar, baz, bob, narf, poink,
> >> poink/narf = 8 ZVals? (That seems logical to me because each its its
> own
> >> variable that just happens to be an array, but I want to be sure.)
> >>
> >> --Larry Garfield
> >>
> >> On Wednesday, January 19, 2011 1:01:44 am Ben Schmidt wrote:
> >>>
> >>> It does the whole of $b. It has to, because when you change 'baz', a
> >>> reference in 'bar' needs to change to point to the newly copied 'baz',
> so
> >>> 'bar' is written...and likewise 'foo' is written.
> >>>
> >>> Ben.
> >>>
> >>> On 19/01/11 5:45 PM, Larry Garfield wrote:
> >>>>
> >>>> Hi folks. I have a question about the PHP runtime that I hope is
> >>>> appropriate for this list. (If not, please thwap me gently; I bruise
> >>>> easily.)
> >>>>
> >>>> I know PHP does copy-on-write. However, how "deeply" does it copy
> when
> >>>> dealing with nested arrays?
> >>>>
> >>>> This is probably easiest to explain with an example...
> >>>>
> >>>> $a['foo']['bar']['baz'] = 1;
> >>>> $a['foo']['bar']['bob'] = 1;
> >>>> $a['foo']['bar']['narf'] = 1;
> >>>> $a['foo']['poink']['narf'] = 1;
> >>>>
> >>>> function test($b) {
> >>>>
> >>>> // Assume each of the following lines in isolation...
> >>>>
> >>>> // Does this copy just the one variable baz, or the full array?
> >>>> $b['foo']['bar']['baz'] = 2;
> >>>>
> >>>> // Does this copy $b, or just $b['foo']['poink']?
> >>>> $b['foo']['poink']['stuff'] = 3;
> >>>>
> >>>> return $b;
> >>>>
> >>>> }
> >>>>
> >>>> // I know this is wasteful; I'm trying to figure out just how
> wasteful.
> >>>> $a = test($a);
> >>>>
> >>>> test() in this case should take $b by reference, but I'm trying to
> >>>> determine how much of a difference it is. (In practice my use case
> has
> >>>> a vastly larger array, so any inefficiencies are multiplied.)
> >>>>
> >>>> --Larry Garfield
> >>
> >
> > --
> > PHP Internals - PHP Runtime Development Mailing List
> > To unsubscribe, visit: http://www.php.net/unsub.php
> >
> >
>
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