Re: [Gumstix-users] Serial Code Question

[Black, M. <Michael.Black@EssexCorp.com>]

My point is that you may NEVER get what you want if for, whatever
reason, the atimega doesn't send the "1" you expect.  It may not send
anything at all (like if the connector falls off).
That's why the timeout loop is what I would do.
Otherwise your program will appear to hang and you won't know why.

You can then decide what you want to do on a timeout -- start all over
again -- or keep trying ad nauseam -- in either case you can print out
an error message on the console to let you know what's happening.

You just need to use a non-blocking read instead of the blocking read.
___________________
Michael D. Black
Essex
bl...@es...

-----Original Message-----
From: gum...@li...
[mailto:gum...@li...] On Behalf Of
Demetris Zavorotnichenko
Sent: Tuesday, June 26, 2007 6:22 AM
To: 'General mailing list for gumstix users.'
Subject: Re: [Gumstix-users] Serial Code Question

I don't really mind if the process sucks all the CPU since I am not
doing
anything else anyway intill the Initialization is complete. So wouldn't
that
be OK if I would do a=20
While=20
{
}

Loop until I get what I want ?

I mean in the future I could just put a "Please" wait window and just
wait
for something to finish.

Wouldn't that work ?


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=20
=20
=20
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=20
=20
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