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From Tom Hobbs <tvho...@googlemail.com>
Subject Re: client hang in com.sun.jini.jeri.internal.mux.Mux.start()
Date Fri, 29 Apr 2011 19:26:39 GMT
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
>

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