"Paul"  skrev i en meddelelse
news:ca236fb1.0405240659.59177c11{at}posting.google.com...
> Hi,
>
> I want to develop two applications that will automatically begin to
> talk to each other via winsock if they are both started up. I want to
> develop it so that no timers are involved in initiating communication.
> The way I thought it would work is that each application would have
> three winsock controls on its form. On the form load event of both
> applications, each would listen out on a different port for incoming
> messages (e.g. A on 4000 and B on 4001). At the same time one of the
> other winsock controls on each form would try to connect to the other
> app on different ports. For example, in app A I might have
>
> wskListen.LocalPort = 4000
> wskListen.Listen
>
> wskSend.RemotePort = 4001
> wskSend.RemoteHost = "127.0.0.1"
> wskSend.Connect
>
> and on app B I'd have
>
> wskListen.LocalPort = 4001
> wskListen.Listen
>
> wskSend.RemotePort = 4000
> wskSend.RemoteHost = "127.0.0.1"
> wskSend.Connect
>
> (The third winsock control on each app is there to accept connection
> requests so isn't important.)
>
> Depending on which app was started first there would always be an
> application listening and when the other was started then there would
> be a connection. On the connect event of that winsock control, I'd
> send a message to the port on which the other application was
> listening asking it to retry connecting. This works okay if I start
> either app A or B first. However if I close one of the applications
> and start it again connection is never reestablished. I get a
> "Connection is aborted due to timeout or other failure" error. I have
> to end both applications in order for them to start communicating
> again. Has anybody managed to achieve something like this.
>

First : Do you know that there is a 2 min "block" for that port you just
used !! Wait 2 minutes and then try again.........I havent found out why,
neither do I have a solution.............

Second : When you disconnect after first run, you must make sure that
it closes all socket's on boot machines, and then starts listen again !!



> Thanks,
>
> Paul
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