ERROR: signing the return value of new by reference

[johnny]

I have the following PHP code. I am using PHP 5.1.6.

$req =& new HTTP_Request($this->accessUrl);

I am getting the following errors:
Strict Standards: Assigning the return value of new by reference is
deprecated in c:\mysite\lib\simpleTest.php on line 80

Can anyone tell why is this? Does it mean assigning the return value
of new by reference is not supported in PHP 5.1.6?

[Andy Hassall]

On 19 Dec 2006 14:27:26 -0800, "johnny" <ra*******@gmail.comwrote:

>I have the following PHP code. I am using PHP 5.1.6.

$req =& new HTTP_Request($this->accessUrl);

I am getting the following errors:
Strict Standards: Assigning the return value of new by reference is
deprecated in c:\mysite\lib\simpleTest.php on line 80

Can anyone tell why is this? Does it mean assigning the return value
of new by reference is not supported in PHP 5.1.6?

Objects are always passed by reference in PHP 5 hence the warning of
deprecation of the effective double-reference.

Obviously if you have to support PHP 4 then you must keep the & - in which
case you need to disable E_STRICT from error_reporting.

--
Andy Hassall :: an**@andyh.co.uk :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

[seaside]

Andy Hassall schrieb:

On 19 Dec 2006 14:27:26 -0800, "johnny" <ra*******@gmail.comwrote:

I have the following PHP code. I am using PHP 5.1.6.

$req =& new HTTP_Request($this->accessUrl);

I am getting the following errors:
Strict Standards: Assigning the return value of new by reference is
deprecated in c:\mysite\lib\simpleTest.php on line 80

Objects are always passed by reference in PHP 5 hence the warning of
deprecation of the effective double-reference.

Obviously if you have to support PHP 4 then you must keep the & - in which
case you need to disable E_STRICT from error_reporting.

Just for the sake of clearness: Any assignment of an object to a
variable creates a copy of the object? A shallow object, I suppose?

[Hendri Kurniawan]

Why would you want to do that...
I mean assigning reference to a newly created
object just doesn't make sense...
$req = new HTTP_Request($this->accessUri);
will just do.

IMHO

johnny wrote:

I have the following PHP code. I am using PHP 5.1.6.

$req =& new HTTP_Request($this->accessUrl);

I am getting the following errors:
Strict Standards: Assigning the return value of new by reference is
deprecated in c:\mysite\lib\simpleTest.php on line 80

Can anyone tell why is this? Does it mean assigning the return value
of new by reference is not supported in PHP 5.1.6?

[Andy Hassall]

On 19 Dec 2006 17:42:29 -0800, "seaside" <se********@mac.comwrote:

>Andy Hassall schrieb:

>On 19 Dec 2006 14:27:26 -0800, "johnny" <ra*******@gmail.comwrote:

>I have the following PHP code. I am using PHP 5.1.6.

$req =& new HTTP_Request($this->accessUrl);

I am getting the following errors:
Strict Standards: Assigning the return value of new by reference is
deprecated in c:\mysite\lib\simpleTest.php on line 80

Objects are always passed by reference in PHP 5 hence the warning of
deprecation of the effective double-reference.

Obviously if you have to support PHP 4 then you must keep the & - in which
case you need to disable E_STRICT from error_reporting.

Just for the sake of clearness: Any assignment of an object to a
variable creates a copy of the object? A shallow object, I suppose?

Assignment with "=" instead of "=&" on PHP4 creates a shallow copy of the
object.

Assignment with "=" on PHP5 does not create a copy; you use the clone keyword
for that, which by default makes a shallow copy, but you can override the
__clone() method in the object to do any deep copying or other operations if
required.

--
Andy Hassall :: an**@andyh.co.uk :: http://www.andyh.co.uk
http://www.andyhsoftware.co.uk/space :: disk and FTP usage analysis tool

[Michael Fesser]

..oO(Hendri Kurniawan)

>Why would you want to do that...
I mean assigning reference to a newly created
object just doesn't make sense...

In PHP 4 it did.

>$req = new HTTP_Request($this->accessUri);
will just do.

This code executed in PHP 4 would have created _two_ objects: the one
created by "new" and a copy of it assigned to $req.

Micha