Re: [Php-java-bridge-users] How to wrap two PHP objects asjavaclosures, and in Java call a method

Hi Jost,

> Do you want to resolve the parameter or the argument?
> I assume you want to unwrap the wrapped PHP object.
> It might be a good idea to extend java_values() this way.
I think in order to avoid confusion between "argument" and "parameter",
I would say I want to "unwrap" the wrapped PHP object.
But:
invokeA() is called with a $Proxy0-object (ID v="4")
How should java_values() be able to "unwrap" the proxy
in order to get the actual PHP object which has been
wrapped before using java_closure() resulting in an
object with ID v="3"?
It is not clear to me how this "ID-mapping" should work!?

> However you can always access your PHP object through other mechanisms.
Any hint how?

What I already tried on the Java-side is to get a handle on the object that has been created with java_closure().
But I could not find out how to come from the $Proxy0-object to the PhpProcedure (which I assume is the one which has ID v="3" which I'm looking for).

Other question regarding Java.inc:
invokeBegin() is defined like this:
  function invokeBegin($object, $method)
but called so:
  $client->protocol->invokeBegin(0, "makeClosure", "5", "6");
What's the meaning of "5" and "6" if invokeBegin() can only accept 2 parameters?

Regards Jürgen

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Von: php...@li... [mailto:php...@li...] 
Gesendet: Samstag, 18. Januar 2009 15:21
An: php...@li...
Betreff: Re: [Php-java-bridge-users] How to wrap two PHP objects asjavaclosures,and in Java call a method on the first closure with the secondclosureas parameter?

> Why shouldn't it be possible to resolve the actual
> parameter $b to the actual PHP-object directly on the
> PHP-side without going back to Java?

Do you want to resolve the parameter or the argument? I assume you want to unwrap the wrapped PHP object. It might be a good idea to extend java_values() this way.

However you can always access your PHP object through other mechanisms.


Regards,
Jost Bökemeier