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ArrayObject - copy or reference

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  ArrayObject - copy or reference Thomas Mlynarczyk <thomas@mlynarczyk-webdesign.de> - 2011-08-08 19:28 +0200
    Re: ArrayObject - copy or reference Jo Schulze <antispam@feuersee.de> - 2011-08-17 00:03 +0200
      Re: ArrayObject - copy or reference Thomas Mlynarczyk <thomas@mlynarczyk-webdesign.de> - 2011-08-18 13:48 +0200
        Re: ArrayObject - copy or reference Jerry Stuckle <jstucklex@attglobal.net> - 2011-08-18 13:54 -0400
          Re: ArrayObject - copy or reference Thomas Mlynarczyk <thomas@mlynarczyk-webdesign.de> - 2011-08-18 20:44 +0200
            Re: ArrayObject - copy or reference Jerry Stuckle <jstucklex@attglobal.net> - 2011-08-18 19:26 -0400

#2754 — ArrayObject - copy or reference

Please consider the following example code:

$array = array( 1, 2, 3 );
$obj = new ArrayObject( $array );
$obj[0] = 42;
echo $array[0];

With PHP 5.2.0 this prints 42, with PHP 5.2.4 however, I get 1 (as 
expected). I have searched both on PHP's changelog [1] and Google, but 
could not find any information about this issue. I assume it is a bug 
that was fixed, but which one and in which version?

Greetings,
Thomas

[1] <http://php.net/ChangeLog-5.php>


-- 
Ce n'est pas parce qu'ils sont nombreux à avoir tort qu'ils ont raison!
(Coluche)

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#2858

Thomas Mlynarczyk wrote:

> Please consider the following example code:
> 
> $array = array( 1, 2, 3 );
> $obj = new ArrayObject( $array );
> $obj[0] = 42;
> echo $array[0];
> 
> With PHP 5.2.0 this prints 42, with PHP 5.2.4 however, I get 1 (as
> expected). I have searched both on PHP's changelog [1] and Google, but
> could not find any information about this issue. I assume it is a bug
> that was fixed, but which one and in which version?

With PHP5.3.7rc this prints 42 which I would expect. The line
$obj[0] = 42;
overwites the value with hash key 0 

I see no bug here.

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#2870

Jo Schulze schrieb:

>> $array = array( 1, 2, 3 );
>> $obj = new ArrayObject( $array );
>> $obj[0] = 42;
>> echo $array[0];
>>
>> With PHP 5.2.0 this prints 42, with PHP 5.2.4 however, I get 1

> With PHP5.3.7rc this prints 42 which I would expect.

No it doesn't. I just installed PHP 5.3.7RC5 to test it. It prints 1.

> The line
> $obj[0] = 42;
> overwites the value with hash key 0 

It overwrites the value with hash key 0 in *$obj*. But it should not 
modify *$array*.

> I see no bug here.

There has definitely been a change in behaviour somewhere between 5.2.0 
and 5.2.4. And that should be documented somewhere.

Greetings,
Thomas

-- 
Ce n'est pas parce qu'ils sont nombreux à avoir tort qu'ils ont raison!
(Coluche)

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#2880

On 8/18/2011 7:48 AM, Thomas Mlynarczyk wrote:
> Jo Schulze schrieb:
>
>>> $array = array( 1, 2, 3 );
>>> $obj = new ArrayObject( $array );
>>> $obj[0] = 42;
>>> echo $array[0];
>>>
>>> With PHP 5.2.0 this prints 42, with PHP 5.2.4 however, I get 1
>
>> With PHP5.3.7rc this prints 42 which I would expect.
>
> No it doesn't. I just installed PHP 5.3.7RC5 to test it. It prints 1.
>
>> The line
>> $obj[0] = 42;
>> overwites the value with hash key 0
>
> It overwrites the value with hash key 0 in *$obj*. But it should not
> modify *$array*.
>
>> I see no bug here.
>
> There has definitely been a change in behaviour somewhere between 5.2.0
> and 5.2.4. And that should be documented somewhere.
>
> Greetings,
> Thomas
>

Thomas,

I get 1 here also (php 5.2.4).  I would consider this a bug - you might 
want to post it to bugs.php.net.

-- 
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

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#2882

Jerry Stuckle schrieb:

> I get 1 here also (php 5.2.4).  I would consider this a bug - you might 
> want to post it to bugs.php.net.

But the bug was in 5.2.0 and seems now fixed in 5.2.4. Please look at my 
code example again. I start with an array $array. Then I use that to 
create an ArrayObject $obj, which should be a *copy* of $array, not a 
reference to it. Now, whatever I do to $obj should not modify $array:

$array = array( 1, 2, 3 );
$obj = new ArrayObject( $array );
$obj[0] = 42;
echo $obj[0]; // This should print 42 and it does
echo '<br>';
echo $array[0]; // This should print 1 and it does in 5.2.4

 From your and Jo's postings I get the impression that you both think 
that $obj should be a reference to $array instead of an independent 
copy. But the docs do not mention any passing by reference (like "&$input"):

ArrayObject::__construct()  ([ mixed $input  [, int $flags  [, string 
$iterator_class  ]]] )

There is, however, a user comment addressing this issue and giving a 
code example almost identical to mine:

<quote>Note that the first argument to ArrayObject::__construct, the 
initial array, is passed by reference. ...</quote>

Considering the date of that comment, it certainly refers to the old 
"pre-5.2.4" behaviour.

Do you think this change in behaviour was unintentional (and thus not 
meant to be a bugfix) and nobody has noticed it so far and that's why I 
cannot find any official documentation about it? Anyway, I think the new 
behaviour is the "correct" one -- I would not expect $array to be passed 
by reference when it's not explicitly documented. Also, this behaviour 
makes more sense to me.

Greetings,
Thomas


-- 
Ce n'est pas parce qu'ils sont nombreux à avoir tort qu'ils ont raison!
(Coluche)

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#2909

On 8/18/2011 2:44 PM, Thomas Mlynarczyk wrote:
> Jerry Stuckle schrieb:
>
>> I get 1 here also (php 5.2.4). I would consider this a bug - you might
>> want to post it to bugs.php.net.
>
> But the bug was in 5.2.0 and seems now fixed in 5.2.4. Please look at my
> code example again. I start with an array $array. Then I use that to
> create an ArrayObject $obj, which should be a *copy* of $array, not a
> reference to it. Now, whatever I do to $obj should not modify $array:
>
> $array = array( 1, 2, 3 );
> $obj = new ArrayObject( $array );
> $obj[0] = 42;
> echo $obj[0]; // This should print 42 and it does
> echo '<br>';
> echo $array[0]; // This should print 1 and it does in 5.2.4
>
>  From your and Jo's postings I get the impression that you both think
> that $obj should be a reference to $array instead of an independent
> copy. But the docs do not mention any passing by reference (like
> "&$input"):
>
> ArrayObject::__construct() ([ mixed $input [, int $flags [, string
> $iterator_class ]]] )
>
> There is, however, a user comment addressing this issue and giving a
> code example almost identical to mine:
>
> <quote>Note that the first argument to ArrayObject::__construct, the
> initial array, is passed by reference. ...</quote>
>
> Considering the date of that comment, it certainly refers to the old
> "pre-5.2.4" behaviour.
>
> Do you think this change in behaviour was unintentional (and thus not
> meant to be a bugfix) and nobody has noticed it so far and that's why I
> cannot find any official documentation about it? Anyway, I think the new
> behaviour is the "correct" one -- I would not expect $array to be passed
> by reference when it's not explicitly documented. Also, this behaviour
> makes more sense to me.
>
> Greetings,
> Thomas
>
>

Yes, I think it's a bug in 5.2.0.  But thinking more of it, I guess they 
won't accept a bug against 5.2.0 any more.

-- 
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
==================

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