Arrays and Pointers

I have come across an issue using arrays ond objects; I am using an
array to keep track of 'n' number of object(The code snippet below is
from my initial testing)

I first declare the object ($nr = new news_resource();), then call a
function of the object ($nr->params_init(); ). This is the put into
the first element of the db, I then call the function again and put the
object in the next element array.

However when I print it all out at the end, all the elements of the
array store the object with the same values...

if I redeclare the object ($nr = new news_resource(); ) again, before
calling the funciton then it gets added in fine.

I'm sure its a typical behaviour and is to do with pointers - but I
don't understand why reassiging the object ($obj = new my_class() )
before callings its funciton should break the link to the original
object (and maintain its state).

In my mind it should either always put a copy in the array or else
always be a pointer.

Can anyone explain why PHP works this way?

Code snippet below:

$nr = new news_resource();

$nr->params_init( 0, "www.num1.com", "Link to num1", "This links to
number 1", 0, 0 );
$resources[] = $nr;

$nr = new news_resource(); //why does this mean that the code works
and the array keeps a copy of the object instead of a pointer?
$nr->params_init( 0, "www.num2.com", "Link to num2", "This links to
number 2", 0, 0 );
$resources[] = $nr;

$nr = new news_resource();
$nr->params_init( 0, "www.num3.com", "Link to num3", "This links to
number 3", 0, 0 );
$resources[] = $nr;



Regards,
Rick Huby
www.e-connected.com

re: Arrays and Pointers

thehuby said the following on 29/09/2005 15:39:[color=blue]
> I have come across an issue using arrays ond objects; I am using an
> array to keep track of 'n' number of object(The code snippet below is
> from my initial testing)
>
> I first declare the object ($nr = new news_resource();), then call a
> function of the object ($nr->params_init(); ). This is the put into
> the first element of the db, I then call the function again and put the
> object in the next element array.
>
> However when I print it all out at the end, all the elements of the
> array store the object with the same values...[/color]

Yes, because $nr is a reference to the news_resource object. Every time
you add it to the array (with $resources[] = $nr), you're inserting
*another* reference to the same object.
i.e.:

$a = new object();
$b = $a;

$b and $a refer to the *same* object. Anything you to do $b will happen
to $a, because they're just aliases of each other.
[color=blue]
> if I redeclare the object ($nr = new news_resource(); ) again, before
> calling the funciton then it gets added in fine.[/color]

In this case, you're redefining $nr as a reference to a new object, so
when you do $resources[] = $nr now, you're storing a reference to a new,
separate object.


--
Oli

re: Arrays and Pointers

thehuby wrote:
[color=blue]
> I have come across an issue using arrays ond objects; I am using an
> array to keep track of 'n' number of object(The code snippet below is
> from my initial testing)
>
> I first declare the object ($nr = new news_resource();), then call a
> function of the object ($nr->params_init(); ). This is the put into
> the first element of the db, I then call the function again and put the
> object in the next element array.
>
> However when I print it all out at the end, all the elements of the
> array store the object with the same values...
>
> if I redeclare the object ($nr = new news_resource(); ) again, before
> calling the funciton then it gets added in fine.
>
> I'm sure its a typical behaviour and is to do with pointers - but I
> don't understand why reassiging the object ($obj = new my_class() )
> before callings its funciton should break the link to the original
> object (and maintain its state).
>
> In my mind it should either always put a copy in the array or else
> always be a pointer.
>
> Can anyone explain why PHP works this way?
>
> Code snippet below:
>
> $nr = new news_resource();
>
> $nr->params_init( 0, "www.num1.com", "Link to num1", "This links to
> number 1", 0, 0 );
> $resources[] = $nr;
>
> $nr = new news_resource(); //why does this mean that the code works
> and the array keeps a copy of the object instead of a pointer?
> $nr->params_init( 0, "www.num2.com", "Link to num2", "This links to
> number 2", 0, 0 );
> $resources[] = $nr;
>
> $nr = new news_resource();
> $nr->params_init( 0, "www.num3.com", "Link to num3", "This links to
> number 3", 0, 0 );
> $resources[] = $nr;[/color]

Odd... Here's what happens for me:
<?php
class myTest{
var $_val=NULL;
var $_mark=NULL;
function init($x){
$this->_val=$x;
}
}

$testar=array();
for($i=0;$i<5;$i++){
$tmp=new myTest();
$tmp->init($i);
$testar[]=$tmp;
}

var_dump($testar);
?>

Output:
array(5) {
[0]=>
object(myTest)#1 (2) {
["_val"]=>
int(0)
["_mark"]=>
NULL
}
[1]=>
object(myTest)#2 (2) {
["_val"]=>
int(1)
["_mark"]=>
NULL
}
[2]=>
object(myTest)#3 (2) {
["_val"]=>
int(2)
["_mark"]=>
NULL
}
[3]=>
object(myTest)#4 (2) {
["_val"]=>
int(3)
["_mark"]=>
NULL
}
[4]=>
object(myTest)#5 (2) {
["_val"]=>
int(4)
["_mark"]=>
NULL
}
}

My thinking on this is that you are somehow using database calls that
are causing this to happen. What version of PHP? Let's see the
news_resource class.

--
Justin Koivisto, ZCE - justin@koivi.com
http://koivi.com

re: Arrays and Pointers

thehuby wrote:
[color=blue]
> I have come across an issue using arrays ond objects; I am using an
> array to keep track of 'n' number of object(The code snippet below is
> from my initial testing)
>
> I first declare the object ($nr = new news_resource();), then call a
> function of the object ($nr->params_init(); ). This is the put into
> the first element of the db, I then call the function again and put the
> object in the next element array.
>
> However when I print it all out at the end, all the elements of the
> array store the object with the same values...[/color]

Oops, I didn't read it close enough...
[color=blue]
> if I redeclare the object ($nr = new news_resource(); ) again, before
> calling the funciton then it gets added in fine.
>
> I'm sure its a typical behaviour and is to do with pointers - but I
> don't understand why reassiging the object ($obj = new my_class() )
> before callings its funciton should break the link to the original
> object (and maintain its state).
>
> In my mind it should either always put a copy in the array or else
> always be a pointer.
>
> Can anyone explain why PHP works this way?
>
> Code snippet below:
>
> $nr = new news_resource();[/color]

Assume:
$nr == 0xffff6224
[color=blue]
> $nr->params_init( 0, "www.num1.com", "Link to num1", "This links to
> number 1", 0, 0 );
> $resources[] = $nr;[/color]

$resources[0] == 0xffff6224
[color=blue]
> $nr = new news_resource(); //why does this mean that the code works
> and the array keeps a copy of the object instead of a pointer?[/color]

If you do the above line:
$nr == 0xffff65a4

Otherwise:
$nr == 0xffff6224
[color=blue]
> $nr->params_init( 0, "www.num2.com", "Link to num2", "This links to
> number 2", 0, 0 );
> $resources[] = $nr;[/color]

If you reset $nr with new:
$nr == 0xffff6c40

else:
$resources[1] == 0xffff6224
[color=blue]
> $nr = new news_resource();
> $nr->params_init( 0, "www.num3.com", "Link to num3", "This links to
> number 3", 0, 0 );
> $resources[] = $nr;[/color]

If you reset $nr with new:
$nr == 0xffff7cc2

else:
$resources[2] == 0xffff6224

So now you have 2 possible outputs:

not resetting $nr each time:
array(
0xffff6224
0xffff6224
0xffff6224
0xffff6224
);

Resetting $nr using "new" each time:
array(
0xffff6224
0xffff65a4
0xffff6c40
0xffff7cc2
);

Look at the above numbers assuming they are memory addresses... do you
see why it behaves that way?

--
Justin Koivisto, ZCE - justin@koivi.com
http://koivi.com

re: Arrays and Pointers

I do see why it behaves that way...my issue is that logically my brain
says it shouldn't. I guess I just don't get why redeclaring the object
creates a second object with a new pointer for the next array element.

I think I was having a moan cos I had spent too much time on this
issue!

Thanks for the info though..it does clear up my understanding.

Regards,

Rick