Progress on preventing multiple instances of a program fromrunning in windows

[Lynch, Jonathan]

The fact that, on my windows XP computer, Rev is automatically setting
itself up to listen to a port to which it just wrote a message seems
awfully odd.

My understanding of ports and sockets is weak - but shouldn't it be able
to write to socket "127.0.0.1:48953" without setting itself up as
listening to port 48953?

Is there a shell command for writing to a socket that could be used that
might not have this behavior?

My method of closing port 48953 after writing to "127.0.0.1:48953" (and
doing the same with port 48954 for communicating in the reverse
direction) works to allow two instances of a stack, running under two
different instances of Rev, to communicate with each other - however,
when I save the same stack as a standalone and try it with two instances
of that standalone, they only manage to pass a couple messages before
getting confused.

If it were possible to write to socket without trying to steal the
listening of that stack, it would be much easier.

-----Original Message-----
From: use-revolution-bounces at lists.runrev.com
[mailto:use-revolution-bounces at lists.runrev.com] On Behalf Of Dar Scott
Sent: Friday, December 03, 2004 12:46 PM
To: How to use Revolution
Subject: Re: Progress on preventing multiple instances of a program
fromrunning in windows


On Dec 3, 2004, at 10:34 AM, Frank D. Engel, Jr. wrote:

> Under Mac OS X 10.3.6, the result is empty if the "accept" command 
> works, and nonempty if it does not work.
>
> If the "accept" command gives a port number which is in use, the 
> "accept" command fails, and the result is nonempty.
>
> That is the behavior I am seeing.

That is what I see on OS X, too.  I think we are seeing a consistent 
pattern with the few comments.  This works as expected on OS X and not 
as expected on Windows.

Dar
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     Dar Scott Consulting
     http://www.swcp.com/dsc/
     Programming Services
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