On Wed, 19 Jan 2011 14:23:49 -0000, Martin Scotta <martinsco...@gmail.com>
wrote:
What about objects?
With objects less copying occurs because the object value (zval) data is
actually just a pointer and an id that for most purposes works as a
pointer.
However, it should be said that while a copy of an array forces more
memory to be copied, the inner zvals are not actually copied. In this
snippet:
$a = array(1, 2, array(3));
$b = $a;
function separate(&$dummy) { }
separate($a);
the copy that occurs when you force the separation of the zval that is
shared by $a and $b ($b = $a doesn't copy the array in $a to $b, it merely
copies the zval pointer of $a to $b and increments its reference count) is
just a shallow copy of hash table and a increment of the first level
zvals' refcounts. This means the zvals that have their pointers stored in
the array $a's HashTable are not themselves copied.
Interestingly (or should I say, unfortunately), this happens even if the
inner zvals are references. See
http://php.net/manual/en/language.references.whatdo.php the part on arrays.
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( $bar );
This example shows no copying (in the sense of "new zval allocation on
passing or assignment") at all.
---
Also... is it better to pass an object as a parameter rather than many
values?
function withValues($anInteger, $aBool, $aString) {
var_dump($anInteger, $aBool, $aString);
}
function withObject(ParamOject $o) {
var_dump( $o->theInteger(), $o->theBool(), $o->theString() );
}
It should be indifferent. In normal circumstances, there is no zval
copying at all (only the pointers of arguments' symbols are copied). Only
when you start throwing references into the mix will you start forcing
copied.
--
Gustavo Lopes
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