Btw, “stacklesslib” contains a class, called util.qchannel which has a built-in queue. it would behave just as you expected. see http://bitbucket.org/krisvale/stacklesslib
K From: [email protected] [mailto:[email protected]] On Behalf Of Lars van Gemerden Sent: 24. maí 2013 17:08 To: The Stackless Python Mailing List Subject: Re: [Stackless] must be something simple Ok, thank you! ==================================== Lars van Gemerden [email protected]<mailto:[email protected]> +31 6 26 88 55 39 ==================================== On 24 mei 2013, at 15:15, Kristján Valur Jónsson <[email protected]<mailto:[email protected]>> wrote: Because channels are not queues. If there is no one waiting to receive, then channel.send() blocks. Channels are “rendezvous points” where one tasklets hands over a piece of data to the other. the “preference” only matters if a transaction can take place. In this case, if a receiver had been blocked on channel, then send would indeed not block. K From: [email protected]<mailto:[email protected]> [mailto:[email protected]] On Behalf Of lars van Gemerden Sent: 24. maí 2013 09:36 To: The Stackless Python Mailing List Subject: Re: [Stackless] must be something simple Hi, I must be mssing something basic; why does the following not print anything: channel = stackless.channel() channel.preference = 1 def run(a): print channel.send(a) print channel.receive() if __name__ == "__main__": stackless.tasklet(run)("anything") stackless.run() If with preference = 1 the send(..) returns immediately, why doesn't "print channel.send(a)" print "None" or something and why is 'a' never received? Cheers, Lars ==================================== Lars van Gemerden [email protected]<mailto:[email protected]> +31 6 26 88 55 39 ==================================== _______________________________________________ Stackless mailing list [email protected]<mailto:[email protected]> http://www.stackless.com/mailman/listinfo/stackless
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