You might want to check out www.winsockvb.com

Max Healey wrote:
> I think that is the key you need to close the port when the client
> closes the connection
>
> This is simple if it is a graceful close that is you use the close
> connection method on the client app.
>
> you can use the connection close event on the server.
>
> but because TCP/IP is essentially a disconnected protocol if the
> client app crashes the server only finds out that when it try's to
> send something to the client when no response is received an error is
> raised so I normally have some kind of a heart beat routine on the
> server just sending one byte every few seconds to ensure the client
> is still there.
>
> another way is to use a control array with the first Winsock control
> listening and passing the connection request to a second new instance
> of the control that way you are always listening for requests.
>
> I have not had any issue with ports being blocked for two minutes.
>
> there is good information at www.vbip.com
>
> also search MSDN on the Winsock control they have a good example of
> the control array version.
>
> Good luck
>
>
> "Steen Gellett"  wrote in message
> news:40b35cd6$0$175$edfadb0f{at}dread11.news.tele.dk...
>>
>> "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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