[Stackless] sendall

Christian Tismer tismer at tismer.com
Thu Oct 10 10:03:05 CEST 2002


Eric van Riet Paap wrote:
> It's the fact that chan.send() is followed by another chan.send(), 
> that's what I mean by hardcoded. What I would prefer to do is send to 
> all current receivers. something like:
> 
> for i in range(chan.numberOfCurrentReceivers):
>     chan.send()

Yep, bt the qustion is *what* you want to send.
This version of send() is meant to send a specific
value to one receiver.
If there is no specific message, just the fact that you
are sending, things are different.

> or
> 
> chan.sendall() #or chan.broadcast()

Broadcast will be implemented, soon.
This will also serve as a notification
mchanism. Broadcasting does not block.
It sends just once the same message to all
receivers in the chain.

> I have looked at the docstrings to see if this information is available 
> but I couldn't find it. I, for a moment, thought channel.queue would be 
> what I was looking for!

The queue contains all senders or receivers which
have not been satisfied yet.
This works with broadcast as well.

At the moment my priorities are different since
I'm porting for IronPort. Shortly after, these
things will be addressed in depth.

thanks for your interest - chris

-- 
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