# Pass by reference (5.1.1 vs 5.1.2)

- Thread: https://externals.io/message/24932
- Original archive: https://news-web.php.net/php.internals/24932
- Started by Rishad Omar on 2006-07-24T15:52:38+00:00
- Messages: 6
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---

## #24932 — Rishad Omar — 2006-07-24T15:52:38+00:00

URL: https://externals.io/message/24932#24932

Hello,

I've upgraded from 5.1.1 to 5.1.2 and discovered the following 
unexpected difference. Best shown by example.

<?
    function getArray(&$arr)
    {
        $arr[] = 12;
    }

    getArray($p = array());
    print_r($p);
?>

In php 5.1.1, the $p is passed correctly as reference and so returns 
with the value 12 in its first element. I regard this as correct behaviour.
In php 5.1.2, the $p is passed by value. So on return, $p is still empty!
This can be fixed as follows:
    $p = array();
    getArray($p);



The following example works, as expected, in both 5.1.1 and 5.1.2.

<?
class A
{
        public $i = 9;
}


function doAmend(A &$a) // & not required because class instance passed 
by reference by default
{
        $a->i += 30;
}

        doAmend($b = new A);
        print_r($b);      // this prints the value of member i as 39 as 
expected.

?

---

## #24934 — Matt Sicker — 2006-07-24T16:12:30+00:00

URL: https://externals.io/message/24932#24934 · In reply to: #24932

> Hello,
>
> I've upgraded from 5.1.1 to 5.1.2 and discovered the following
> unexpected difference. Best shown by example.
>
> <?
>     function getArray(&$arr)
>     {
>         $arr[] = 12;
>     }
>
>     getArray($p = array());
>     print_r($p);
> ?>
>
> In php 5.1.1, the $p is passed correctly as reference and so returns
> with the value 12 in its first element. I regard this as correct
> behaviour. In php 5.1.2, the $p is passed by value. So on return, $p
> is still empty! This can be fixed as follows:
>     $p = array();
>     getArray($p);
>
>
>
> The following example works, as expected, in both 5.1.1 and 5.1.2.
>
> <?
> class A
> {
>         public $i = 9;
> }
>
>
> function doAmend(A &$a) // & not required because class instance
> passed by reference by default
> {
>         $a->i += 30;
> }
>
>         doAmend($b = new A);
>         print_r($b);      // this prints the value of member i as 39
> as expected.
>
> ?>
I'm guessing it has something to do with a memory leak.  Either that, or 
scope; creating a variable inside of a function call probably goes out 
of scope after the function returns.  Enable error_reporting(E_ALL|
E_STRICT) and see what happens.
-- 
Matt Sicker

---

## #24939 — Johannes Schlueter — 2006-07-24T17:16:26+00:00

URL: https://externals.io/message/24932#24939 · In reply to: #24932

Hi,


>     getArray($p = array());

Here you are relying on undefined behavior. It is not defined wether $p = 
array() or the function call getarray($p) should be executed first so the 
order might always change.

You should always use the two lines
     $p = array();
     getArray($p);
to be safe.

johannes

---

## #24941 — Todd Ruth — 2006-07-24T17:53:27+00:00

URL: https://externals.io/message/24932#24941 · In reply to: #24939

It looks like
http://www.php.net/manual/en/language.operators.assignment.php
could use a bit of updating.  A paragraph about php 5 or a
link to the semantics changes for php 5 would help newbies.
The text about php4 says there is copy on assignment, which
implies a performance hit for using = instead of =&, but I
believe "copy on change" is more accurate.  It wouldn't hurt
to explicitly say that "=" is generally better than "=&"
(unless you really need an alias).  A user comment has been
there unchallenged for over 2 years saying you can pass an
assignment as a reference to a function, so it isn't surprising
that someone considered the 5.1.1 to 5.1.2 change an issue.
(Not that a user comment should bind the php devs, but when
the comment goes unchallenged, it can lead to false impressions.)
Perhaps some text explaining the assignment returns an r-value
would be helpful.  (If people understand "($x = 1) = 2;"
shouldn't work, they might understand that passing "$x=1" to
a function by reference shouldn't affect $x.  (or should
"($x = 1) = 2;" work and 5.1.2 had a real bug?))

I have to say the documentation for php is completely awesome
and is a major factor in our company using the language.  I
just point out the above to help; it isn't a complaint.

BTW, I disagree with the post below that indicates the issue is
the order of calling.  There is no way to call a function
before doing the work in the argument list.  If I call f(1+2),
I can be sure the computation to get 3 will be performed
_before_ calling f.  The issue with f($p=array()) is that
"$p=array()" is an r-value.

- Todd


> Hi,
> 
> 
> > […]
> 
> Here you are relying on undefined behavior. It is not defined wether $p = 
> array() or the function call getarray($p) should be executed first so the 
> order might always change.
> 
> You should always use the two lines
>      $p = array();
>      getArray($p);
> to be safe.
> 
> johannes
> 
-- 
Todd Ruth <truth@proposaltech.com

---

## #24947 — Ron Korving — 2006-07-24T19:02:40+00:00

URL: https://externals.io/message/24932#24947 · In reply to: #24939

in that case, the documentation on `socket_select()` needs to be edited:

http://www.php.net/socket-select says:

<?php
/* Prepare the read array */
$read = array($socket1, $socket2);

$num_changed_sockets = socket_select($read, $write = NULL, $except = NULL,
0);

if ($num_changed_sockets === false) {
   /* Error handling */
} else if ($num_changed_sockets > 0) {
   /* At least at one of the sockets something interesting happened */
}
?>

according to you, $write and $except could be undefined, right?

- Ron


"Johannes Schlueter" <johannes@php.net> wrote in message
news:200607241916.27024.johannes@php.net...
> Hi,
>
> 
> > […]
>
> Here you are relying on undefined behavior. It is not defined wether $p =
> array() or the function call getarray($p) should be executed first so the
> order might always change.
>
> You should always use the two lines
>      $p = array();
>      getArray($p);
> to be safe.
>
> johannes

---

## #24948 — Ron Korving — 2006-07-24T19:05:16+00:00

URL: https://externals.io/message/24932#24948 · In reply to: #24941

Actually Todd, ...

> There is no way to call a function
> before doing the work in the argument list.  If I call f(1+2),
> I can be sure the computation to get 3 will be performed
> _before_ calling f.  The issue with f($p=array()) is that
> "$p=array()" is an r-value.

...that is theoretically very possible, and it's something called "lazy
evaluation": http://en.wikipedia.org/wiki/Lazy_evaluation

- Ron


"Todd Ruth" <truth@proposaltech.com> wrote in message
news:1153763607.30382.184.camel@notebook.local...
> It looks like
> http://www.php.net/manual/en/language.operators.assignment.php
> could use a bit of updating.  A paragraph about php 5 or a
> link to the semantics changes for php 5 would help newbies.
> The text about php4 says there is copy on assignment, which
> implies a performance hit for using = instead of =&, but I
> believe "copy on change" is more accurate.  It wouldn't hurt
> to explicitly say that "=" is generally better than "=&"
> (unless you really need an alias).  A user comment has been
> there unchallenged for over 2 years saying you can pass an
> assignment as a reference to a function, so it isn't surprising
> that someone considered the 5.1.1 to 5.1.2 change an issue.
> (Not that a user comment should bind the php devs, but when
> the comment goes unchallenged, it can lead to false impressions.)
> Perhaps some text explaining the assignment returns an r-value
> would be helpful.  (If people understand "($x = 1) = 2;"
> shouldn't work, they might understand that passing "$x=1" to
> a function by reference shouldn't affect $x.  (or should
> "($x = 1) = 2;" work and 5.1.2 had a real bug?))
>
> I have to say the documentation for php is completely awesome
> and is a major factor in our company using the language.  I
> just point out the above to help; it isn't a complaint.
>
> BTW, I disagree with the post below that indicates the issue is
> the order of calling.  There is no way to call a function
> before doing the work in the argument list.  If I call f(1+2),
> I can be sure the computation to get 3 will be performed
> _before_ calling f.  The issue with f($p=array()) is that
> "$p=array()" is an r-value.
>
> - Todd
>
> 
> > […]
=
> > […]
the
> > […]
> -- 
> Todd Ruth <truth@proposaltech.com