[Stackless] sendall

[Eric van Riet Paap]

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()

or

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

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!

kind regards,
Eric


On Wednesday, October 9, 2002, at 10:15 PM, Christian Tismer wrote:

> Eric van Riet Paap wrote:
>> Hi,
>> in test/chantest.py it is hardcoded that there are two tasks 
>> listening to channel named 'chan' (two chan.send(...) in a row). What 
>> would be the easiest/best way to remove the hardcoding in this case?
>
> I don't understand what your problem is.
> What is hardcoded?
> The name "chan" happens to be the same in the receiver function
> and in the main, but that's just my laziness.
> What hardcoding would you like to remove?
>
> ciao - chris
>
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