On 8/26/11 4:55 PM, Alan D. Cabrera wrote:
On Aug 26, 2011, at 7:47 AM, Emmanuel Lecharny wrote:
On 8/26/11 4:28 PM, Alan D. Cabrera wrote:
On Aug 26, 2011, at 7:12 AM, Julien Vermillard wrote:
On Fri, Aug 26, 2011 at 4:07 PM, Emmanuel Lecharny<elecha...@gmail.com> wrote:
On 8/26/11 3:44 PM, Alan D. Cabrera wrote:
On Aug 26, 2011, at 6:40 AM, Julien Vermillard wrote:
On Fri, Aug 26, 2011 at 3:24 PM, Alan D. Cabrera<l...@toolazydogs.com>
wrote:
On Aug 26, 2011, at 4:14 AM, Julien Vermillard wrote:
I modified the API to remove IoFilterChain. Now you are supposed to
give a list of filter to the service before starting it :
// create the fitler chain for this service
List<IoFilter> filters = new ArrayList<IoFilter>();
filters.add(new LoggingFilter("byte log filter"));
filters.add(new MyCodecFilter());
filters.add(new LoggingFilter("pojo log filter"));
filters.add(newMyProtocolLogicFilter());
acceptor.setFilters(filters);
acceptor.bind(...);
How do we make chains where two filters feed into one or one filter
feeds two filters? If you look in my sandbox we can accommodate this via:
static import a.m.util.Util. linkParentWithChild; // to be written
IoFilter foo = new FooFilter();
LinkStateFilter link = new LinkStateFilter();
IoFilter checksum = new ChecksumFilter();
IoFilter log = new LogFilter();
link.addLinkStateListener(foo);
linkParentWithChild(foo, checksum);
linkParentWithChild(link, checksum);
linkParentWithChild(checksum, log);
acceptor.setFilters(foo);
About the code in the sandbox :
http://svn.apache.org/repos/asf/mina/sandbox/adc/ahc/mina3/src/main/java/org/apache/mina/core/IoFilter.java
I see no IoFilter.addLinkStateListener(..) method, am I looking at the
right place ?
Oops, it was meant to just be a sketch. :)
About the "filters feed into one or one filter feeds two filters", do
you have a concrete use case in mind for that ?
The above example does, Foo and the link state filter. I'm sure that
we've discussed this before. Another example is a mux/demux situation. How
would all of this fit into the grand scheme of things?
Yeah, it really should be a graph of filters, not a list of filters.
Well if it's just for demuxing I proposed few mails ago this solution
: http://s.apache.org/A9W
I think we need to make graphing a 1st class citizen and not buried inside
another filter class.
I do agree. The proposed solution on http://s.apache.org/A9W is what we
currently have, and it's tedious to manage.
It would be way better to be able to let the controler call the next filter
based on an evaluation method based on the current state.
Now, the question is : how do the controller knows which filter to call next ?
It must have the current state, and it must be able to make the connection.
For instance, let's say that from filter F we can switch to either filter G or
filter H, depending on the state we transit to in F.
F --(state1)--> G
or
F --(state2)--> H
That means the controller has a map { (F, state1, G), (F, state2, H)}
somewhere. State should be passed to the controller too :
...
controller.nextFilter( newState );
...
Pretty theorical at this point... I'm sorry not to have a lot of time to code
this, I do realize that for you guys implementing ideas it's a PITA...
I was thinking of a simple DAG some, or all, of the nodes can be FSMs.
-> [FSM1] -> [Filter] -> [FSM2] -> [Filter]
Inside the FSM there could be chains, but there would be one chain per state.
Not sure I grok what you say here. There is more than one state, and
from each state, you may have more than one transition.
Maybe it's just a problem of vocabulary...
<theory>
In a FSM, you transit from Si to Sj, following a transition Ta, which
depends on a context. You may also transit to a state Sk, following a
different transition Tb, if the context is different.
Selection the transition to follow is all about knowing what's the
context is, and in my sample, this was what I call the 'state', which
was most certainly an error, as it's clearly a transition. I should have
wrote :
F --(transition1)--> G
or
F --(transition2)--> H
where F, G, H are filters (ore "states")
are we on the same page ?
</theory>
--
Regards,
Cordialement,
Emmanuel Lécharny
www.iktek.com
