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Date: Thu, 20 Jan 2011 09:45:14 +1100
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Subject: Re: [PHP-DEV] How deep is copy on write?
From: mail_ben_schmidt@yahoo.com.au (Ben Schmidt)

On 20/01/11 6:05 AM, larry@garfieldtech.com wrote:
> 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?

Yes.

> Related: What is the overhead of a ZVal? I'm assuming it's a fixed
> number of bytes.

It seems not, though a zval has a fixed size. What that size is will
depend on the compiler and architecture of the system being used, or at
least on the ABI.

 From zend.h:

typedef union _zvalue_value {
	long lval;					/* long value */
	double dval;				/* double value */
	struct {
		char *val;
		int len;
	} str;
	HashTable *ht;				/* hash table value */
	zend_object_value obj;
} zvalue_value;

struct _zval_struct {
	/* Variable information */
	zvalue_value value;		/* value */
	zend_uint refcount__gc;
	zend_uchar type;	/* active type */
	zend_uchar is_ref__gc;
};

The zvalue_value union will probably be 8 or 12 bytes, depending on the
architecture. The whole struct will then probably be between 14 and 24
bytes, depending on the architecture and structure alignment and so on.

For my system:

$ cd php-5.3.3
$ ./configure
$ cd Zend
$ gcc -I. -I../TSRM -x c - <<END
> #include "zend.h"
> int main(void) {
>    printf("%lu\n",sizeof(zval));
>    return 0;
> }
> END
$ file ./a.out
./a.out: Mach-O 64-bit executable
$ ./a.out
24
$ gcc -I. -I../TSRM -arch i386 -x c - <<END
> #include "zend.h"
> int main(void) {
>    printf("%lu\n",sizeof(zval));
>    return 0;
> }
> END
$ file ./a.out
./a.out: Mach-O executable i386
$ ./a.out
16

You can figure out what you think the overhead is from that. For a
string, arguably the whole structure is overhead, since the string is
stored elsewhere via pointer. Likewise for objects. For a double, the
payload is 8 bytes, and stored in the zval, so there's less overhead. An
integer, with a payload of 4 bytes, is somewhere in between.

Ben.



>
> --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.
>>
>>
>