Re: Model performance question
The wiki says (https://cwiki.apache.org/WICKET/working-with-wicket-models.html): Compound models allow containers to share models with their children. This saves memory, but more importantly, it makes replication of models much cheaper in a clustered environment. I think what we're saying in this thread is that PropertyModels actually have the same benefits as CPM; that is, a CPM doesn't give you any performance benefits over using a PropertyModel -- there's less typing with a CPM, but it ends up the same in the end. Does that sound right? Andrew On Tue, Jun 25, 2013 at 12:02 PM, Bas Gooren b...@iswd.nl wrote: Hi, It sounds like you know what you are doing, but I just want to check why you chose myObject as a variable name in your example? If you are properly using detachable models and do not want to serialize a large object graph, myObject needs to be a loadable detachable model, and not an actual object in your example. myLDM or myObjectModel would be a more logical name. In my experience it doesn't matter where you create your models, because using a compound property model will automatically create propertymodels on your nested components anyway. Each component needs its own model anyway, to get and set its value. Met vriendelijke groet, Kind regards, Bas Gooren Op 25-6-2013 16:20, schreef gmparker2000: Considering two alternative ways to set a model: ... final CompoundPropertyModel myModel = new CompoundPropertyModel(myObject); control1.setModel(myModel.bind(field1)); control2.setModel(myModel.bind(field2)); control3.setModel(myModel.bind(field3)); ... and ... control1.setModel(new PropertyModel(myObject, field1)); control2.setModel(new PropertyModel(myObject, field2)); control3.setModel(new PropertyModel(myObject, field3)); ... are there any performance benefits of one over the other? I profiled each and they appear equivalent from the number of objects created point of view. I just want to make sure that option two isn't doing something like serializing myObject for each control. It doesn't appear that this is happening but wanted to make sure. We are binding controls to fairly large nested Java Objects. Other than making sure to use LoadableDetachable models where possible are there any other strategies for making sure performance is optimal? For example, would setting the model on the form rather than on each control have any performance benefit? Thanks -- View this message in context: http://apache-wicket.1842946.n4.nabble.com/Model-performance-question-tp4659771.html Sent from the Users forum mailing list archive at Nabble.com. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: Model performance question
Hi, On Tue, Jun 25, 2013 at 5:20 PM, gmparker2000 greg.par...@brovada.comwrote: Considering two alternative ways to set a model: ... final CompoundPropertyModel myModel = new CompoundPropertyModel(myObject); control1.setModel(myModel.bind(field1)); control2.setModel(myModel.bind(field2)); control3.setModel(myModel.bind(field3)); ... and ... control1.setModel(new PropertyModel(myObject, field1)); control2.setModel(new PropertyModel(myObject, field2)); control3.setModel(new PropertyModel(myObject, field3)); ... are there any performance benefits of one over the other? I profiled each and they appear equivalent from the number of objects created point of view. I just want to make sure that option two isn't doing something like serializing myObject for each control. It doesn't appear that this is happening but wanted to make sure. Java serialization writes a given instance just once and all following writes of the same instance will write just pointers to the first write We are binding controls to fairly large nested Java Objects. Other than making sure to use LoadableDetachable models where possible are there any other strategies for making sure performance is optimal? For example, would setting the model on the form rather than on each control have any performance benefit? Thanks -- View this message in context: http://apache-wicket.1842946.n4.nabble.com/Model-performance-question-tp4659771.html Sent from the Users forum mailing list archive at Nabble.com. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
Re: Model performance question
Hi, It sounds like you know what you are doing, but I just want to check why you chose myObject as a variable name in your example? If you are properly using detachable models and do not want to serialize a large object graph, myObject needs to be a loadable detachable model, and not an actual object in your example. myLDM or myObjectModel would be a more logical name. In my experience it doesn't matter where you create your models, because using a compound property model will automatically create propertymodels on your nested components anyway. Each component needs its own model anyway, to get and set its value. Met vriendelijke groet, Kind regards, Bas Gooren Op 25-6-2013 16:20, schreef gmparker2000: Considering two alternative ways to set a model: ... final CompoundPropertyModel myModel = new CompoundPropertyModel(myObject); control1.setModel(myModel.bind(field1)); control2.setModel(myModel.bind(field2)); control3.setModel(myModel.bind(field3)); ... and ... control1.setModel(new PropertyModel(myObject, field1)); control2.setModel(new PropertyModel(myObject, field2)); control3.setModel(new PropertyModel(myObject, field3)); ... are there any performance benefits of one over the other? I profiled each and they appear equivalent from the number of objects created point of view. I just want to make sure that option two isn't doing something like serializing myObject for each control. It doesn't appear that this is happening but wanted to make sure. We are binding controls to fairly large nested Java Objects. Other than making sure to use LoadableDetachable models where possible are there any other strategies for making sure performance is optimal? For example, would setting the model on the form rather than on each control have any performance benefit? Thanks -- View this message in context: http://apache-wicket.1842946.n4.nabble.com/Model-performance-question-tp4659771.html Sent from the Users forum mailing list archive at Nabble.com. - To unsubscribe, e-mail: users-unsubscr...@wicket.apache.org For additional commands, e-mail: users-h...@wicket.apache.org
