On Wed, 2008-09-03 at 18:17 +0200, Lupus Michaelis wrote:
> Robert Cummings a écrit :
>
> > Please explain how it can take up more than twice.
>
> It is obvious for a C developper.
>
> == 8< == Zend/zend.h
> typedef struct _zval_struct zval;
>
> 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;
>
> typedef struct _zend_object {
> zend_class_entry *ce;
> HashTable *properties;
> HashTable *guards; /* protects from __get/__set ... recursion */
> } zend_object;
>
> struct _zval_struct {
> /* Variable information */
> zvalue_value value; /* value */
> zend_uint refcount__gc;
> zend_uchar type; /* active type */
> zend_uchar is_ref__gc;
> };
>
> == 8< ==
>
> When you write
> $a = 1 ;
>
> The $a variable don't contain only the integer, it contains a lot of
> more stuff. This stuff is needed by the fact that a variable can contain
> every types available in PHP.
>
> So, I let you calculate the memory needed to store the variable, but
> you can easily understand that they'll not be twice the previous memory
> footprint, but a little more.
I do develop in C so I now need to take a stick to you. It's still
double space. Use a simple example for yourself. Let's say a struct like
following:
struct _foo
{
int i;
int j;
int k[5];
} foo;
In 32 bit system we have:
32 bits for i
+ 32 bits for j
+ (32 bits for each k) * 5
+ 32 bits for the pointer to foo
= 32 * 8
In 64 bit system we have:
64 bits for i
+ 64 bits for j
+ (64 bits for each k) * 5
+ 64 bits for the pointer to foo
= 64 * 8
= (32 * 2) * 8
= (32 * 8) * 2
Exactly double. Please explain where I went wrong.
Cheers,
Rob.
--
http://www.interjinn.com
Application and Templating Framework for PHP
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
