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From Gregg Wonderly <ge...@cox.net>
Subject Re: client hang in com.sun.jini.jeri.internal.mux.Mux.start()
Date Sat, 30 Apr 2011 15:02:31 GMT
In the history of original Java threading model, and the NIO development to use 
"select/poll" from your own thread, rather than registering call back methods 
(via an interface) kept a lot of development from using a model where threading 
was managed internally by the package or by the JVM.  As a result, we have 
structures like today where notifications are less common.  In this code though, 
I think the structure is internal enough that it's not necessary to really use 
Future or some other mechanisms.

Timeouts are always a "hard way" to manage "loss of functionality" because you 
really don't know when things are "not working", only that something is taking 
longer than your timeout accounted for.   Timeouts can make it possible for more 
pending work to pile up on the other end that might slow the results even more. 
  E.g. you wait 30seconds and retry while the actual work on the other end is 
taking 35 seconds to get to and thus the queue rate exceeds the dequeue rate and 
things start piling up.

If you are going to use a timeout, we need to have some sort of indication from 
both ends perspective that the attempt has been aborted, as early as possible. 
For cases where I/O traffic is written/read, that usually means closing the 
socket.  I'm not sure of the ramifications of doing that, since I haven't looked 
too hard at this code.

Gregg Wonderly

On 4/29/2011 4:38 PM, Mike McGrady wrote:
> Throwing my two cents in here just to state my opinion.  This is an effort that
 > could pay dividends if it were done with a view toward the future - absolutely
 > no pun intended.  I do not know the details of the code but if futures could be
 > useful here, they would be welcomed by myself.
>
> MG
>
>
> On Apr 29, 2011, at 1:33 PM, Christopher Dolan wrote:
>
>> Thanks, Tom.
>>
>> I don't really understand your Future suggestion.  Are you suggesting
>> changing the async handshake to a Future? If so, that sounds like a very
>> involved change, touching a lot of code in Mux and its subclasses.
>>
>> setDown() changes the value of the muxDown boolean to true, so it's a
>> valid way out of the loop.
>>
>> Chris
>>
>> -----Original Message-----
>> From: Tom Hobbs [mailto:tvhobbs@googlemail.com]
>> Sent: Friday, April 29, 2011 2:27 PM
>> To: dev@river.apache.org
>> Subject: Re: client hang in com.sun.jini.jeri.internal.mux.Mux.start()
>>
>> The proposed code looks fine to me.  Two points leap out, more
>> discussion
>> points than anything else.
>>
>> For some reason I've recently developed an aversion to writing my on
>> timeout
>> logic.  did you consider using something like a Future here or might
>> that be
>> serious overkill (it wouldn't surprise me if it was)?
>>
>> Also is Setdown intended to break out of the while loop?  Because I
>> can't
>> see a way to escape it.  (I don't have the rest of the code in front of
>> me)
>>
>> Thanks for keep raising these issues - especially because you usually
>> supply
>> fixes for them!
>>
>> Tom
>>
>> On 29 Apr 2011 19:41, "Christopher Dolan"<christopher.dolan@avid.com>
>> wrote:
>>> I've experienced occasional cases where clients get stuck in the
>>> following block of code in Mux.start. Has anyone experienced this
>>> problem? I have a proposed solution below. Has anyone thought about a
>>> similar solution already?
>>>
>>> -- Current code --
>>> 1 asyncSendClientConnectionHeader();
>>> 2 synchronized (muxLock) {
>>> 3 while (!muxDown&&  !clientConnectionReady) {
>>> 4 try {
>>> 5 muxLock.wait(); // REMIND: timeout?
>>> 6 } catch (InterruptedException e) {
>>> 7 ...
>>> 8 }
>>> 9 }
>>> 10 if (muxDown) {
>>> 11 IOException ioe = new IOException(muxDownMessage);
>>> 12 ioe.initCause(muxDownCause);
>>> 13 throw ioe;
>>> 14 }
>>> 15 }
>>>
>>> -- Explanation of the code --
>>> This code handles the initial client-server handshake that starts a
>> JERI
>>> connection. In line 1, the client sends its 8-byte greeting to the
>>> server. Then in the loop on lines 3-9, it waits for the server's
>>> response. If the reader thread gets a satisfactory response from the
>>> server, it sets clientConnectionReady=true and calls
>>> muxLock.notifyAll(). In all other cases (aborted connection,
>> mismatched
>>> protocol version, etc) the reader invokes Mux.setDown() which sets
>>> muxDown=true and calls muxLock.notifyAll(). In lines 10-14, it throws
>> if
>>> the handshake was a failure.
>>>
>>> In my scenario (which uses simple TCP sockets, nothing fancy), the
>>> invoker thread sits on line 5 indefinitely. My problem hard to
>>> reproduce, so I haven't found out what the server is doing in this
>> case.
>>> I hope to figure that out eventually, but presently I'm interested in
>>> the "REMIND: timeout?" comment.
>>>
>>> -- Timeout solution --
>>> It seems obvious to me that there should be a timeout here. There are
>>> lots of imaginable cases where the client could get stuck here:
>>> server-side deadlock, abrupt server crash, logic error in client Mux
>>> code. You'd expect that the server would either respond with its
>> 8-byte
>>> handshake very quickly or never, so a modest timeout (like 15 or 30
>>> seconds) should be good. If that timeout is triggered, I would expect
>>> that the code above would call Mux.setDown() and throw an IOException.
>>> That exception would either cause a retry or be thrown up to the
>> invoker
>>> as a RemoteException.
>>>
>>> -- Proposed code (untested) --
>>> 3 long now = System.currentTimeMillis();
>>> 4 long endTime = now + timeoutMillis;
>>> 5 while (!muxDown&&  !clientConnectionReady) {
>>> 6 if (now>= endTime) {
>>> 7 setDown("timeout waiting for server to respond
>>> to handshake", null);
>>> 8 } else {
>>> 9 try {
>>> 10 muxLock.wait(endTime - now);
>>> 11 now = System.currentTimeMillis();
>>> 12 } catch (InterruptedException e) {
>>> 13 setDown("interrupt waiting for connection
>>> header", e);
>>> 14 }
>>> 15 }
>>> 16 }
>>>
>>> This code assumes a configurable timeoutMillis parameter has been set
>>> earlier.
>>>
>>> I can't think of any alternative solutions. Putting the timeout in the
>>> Reader logic seems higher risk. There's incomplete code in JERI to
>>> implement a ping packet (see Mux.asyncSendPing, never used), but that
>>> would only be relevant after the initial handshake and wouldn't help
>>> here.
>>>
>>> Thanks,
>>> Chris
>>>
>
> Michael McGrady
> Chief Systems Architect
> Topia Technology, Inc.
> Cel 1.253.720.3365
> Work 1.253.572.9712 extension 2037
> Skype ID: michael.mcgrady5
> mmcgrady@topiatechnology.com
>
>
>
>


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