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Date: Wed, 19 Jan 2011 13:05:23 -0600
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Subject: Re: [PHP-DEV] How deep is copy on write?
From: larry@garfieldtech.com ("larry@garfieldtech.com")

So it sounds like the general answer is that if you pass a complex array 
to a function by value and mess with it, data is duplicated for every 
item you modify and its direct ancestors up to the root variable but not 
for the rest of the tree.

For objects, because of their "pass by handle"-type behavior you are 
(usually) modifying the same data directly so there's no duplication.

Does that sound correct?

Related: What is the overhead of a ZVal?  I'm assuming it's a fixed 
number of bytes.

--Larry Garfield

On 1/19/11 11:27 AM, Gustavo Lopes wrote:
> On Wed, 19 Jan 2011 14:23:49 -0000, Martin Scotta
> <martinscotta@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.
>
>