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Re: Why does Parser.php pass a value to my extension instead of a reference?

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mickewiki at gmail

Oct 10, 2009, 1:25 AM

Post #1 of 3 (697 views)
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Re: Why does Parser.php pass a value to my extension instead of a reference?

Robert Rohde <rarohde <at> gmail.com> writes:

> PHP 4 and PHP 5 handle objects as function parameters differently.
> Mediawiki now considers PHP 5.0 or later as a prerequisite (5.1 or
> later recommended). The warning you quote is a plausible consequence
> of using PHP 4 under certain circumstances, so that is the first thing
> I would check.

Right, I'll see if I can get a hold of the guy again. Thanks for that.

Is there a notable difference between passing the parser object by reference
rather than as a value in this case though? I mean, if I can support PHP4 with
my extension with out any (notable) differences I could just as easily pass the
object as a value instead. I don't notice an difference in performance when I
try it...

/Micke

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innocentkiller at gmail

Oct 10, 2009, 1:31 AM

Post #2 of 3 (645 views)
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Re: Why does Parser.php pass a value to my extension instead of a reference? [In reply to]

On Sat, Oct 10, 2009 at 4:25 AM, Micke Nordin <mickewiki [at] gmail> wrote:
> Robert Rohde <rarohde <at> gmail.com> writes:
>
>> PHP 4 and PHP 5 handle objects as function parameters differently.
>> Mediawiki now considers PHP 5.0 or later as a prerequisite (5.1 or
>> later recommended). The warning you quote is a plausible consequence
>> of using PHP 4 under certain circumstances, so that is the first thing
>> I would check.
>
> Right, I'll see if I can get a hold of the guy again. Thanks for that.
>
> Is there a notable difference between passing the parser object by reference
> rather than as a value in this case though? I mean, if I can support PHP4 with
> my extension with out any (notable) differences I could just as easily pass the
> object as a value instead. I don't notice an difference in performance when I
> try it...
>
> /Micke
>
>
>
>
> _______________________________________________
> Wikitech-l mailing list
> Wikitech-l [at] lists
> https://lists.wikimedia.org/mailman/listinfo/wikitech-l
>

There's no point in targeting PHP4. Mediawiki doesn't support it anymore.

-Chad

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Platonides at gmail

Oct 10, 2009, 2:48 PM

Post #3 of 3 (634 views)
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Re: Why does Parser.php pass a value to my extension instead of a reference? [In reply to]

Micke Nordin writes:
> Robert Rohde <rarohde <at> gmail.com> writes:
>
>> PHP 4 and PHP 5 handle objects as function parameters differently.
>> Mediawiki now considers PHP 5.0 or later as a prerequisite (5.1 or
>> later recommended). The warning you quote is a plausible consequence
>> of using PHP 4 under certain circumstances, so that is the first thing
>> I would check.
>
> Right, I'll see if I can get a hold of the guy again. Thanks for that.
>
> Is there a notable difference between passing the parser object by reference
> rather than as a value in this case though? I mean, if I can support PHP4 with
> my extension with out any (notable) differences I could just as easily pass the
> object as a value instead. I don't notice an difference in performance when I
> try it...
>
> /Micke

You can't support PHP5 in mediawiki.
The warning you get was added on PHP 5.3 and doesn't affect that code
works OK. You should remove the reference to make the warning go away.

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