Yeah, but note that there are two shreds running this loop in parallell. So if one event on each port arrive at the same time, chances are that the first yield will hand over execution to the second shred, that in turn overwrites b.value and b.message. I think you need at least two Bang instances to be sure that this doesn't happen.

/Stefan

On Mon, Apr 28, 2008 at 5:10 PM, Kassen <signal.automatique@gmail.com> wrote:
2008/4/28 Stefan Blixt <stefan.blixt@gmail.com>:

I don't know about the segv signal, but it seems to me that there is only one Bang instance that is shared by all iterations/shreds. This means that if two events arrive at this loop:


       while( oe.nextMsg() ) {
           oe.getInt() => b.value;
           osctype => b.message;
           b.broadcast();
       }

the second's values will overwrite those of the first (value and message from the first event will be lost).

I think that's right and I think the way around this would be


       while( oe.nextMsg() ) {
           oe.getInt() => b.value;
           osctype => b.message;
           b.broadcast();
           //yield to receiving shreds,
           //then continue processing the message cue
           me.yield();
       }

This is the exact same principle I talked about earlier in this topic; if you don't yield you don't give the waiting shreds a chance to run. Advancing time here would probably not be a good idea.

Yours,
Kas.

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